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Converse and contrapositive statements

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55596210
For the conditional statement “If a quadrilateral is a square, then its diagonals are congruent,” identify the hypothesis and the conclusion.

Hints

- Locate the “if” clause first. - Then identify what the statement says must follow from that condition.

Solution

1. The hypothesis is the part following “if.” 2. The conclusion is the part following “then.”

Answer

Hypothesis: the quadrilateral is a square. Conclusion: its diagonals are congruent.
55596310
The statement is: “If two coplanar lines are parallel, then corresponding angles formed by a transversal are congruent.” Which statement is its contrapositive? A. If corresponding angles are congruent, then the lines are parallel. B. If the lines are not parallel, then corresponding angles are not congruent. C. If corresponding angles are not congruent, then the lines are not parallel.

Hints

- A contrapositive reverses the two parts and negates both. - Do not confuse it with the converse, which only reverses the parts.

Solution

1. Write the original form as “If \(P\), then \(Q\).” 2. The contrapositive is “If not \(Q\), then not \(P\).” 3. Here, not \(Q\) is “corresponding angles are not congruent,” and not \(P\) is “the lines are not parallel.”

Answer

C
55596410
The original statement is: “If two angles are vertical angles, then they are congruent.” A second statement says: “If two angles are not vertical angles, then they are not congruent.” Is the second statement the converse, inverse, or contrapositive of the original?

Hints

- Track whether the hypothesis and conclusion were reversed. - Track whether each part was negated.

Solution

1. Write the original as “If \(P\), then \(Q\).” 2. The second statement has the form “If not \(P\), then not \(Q\).” 3. That form is the inverse.

Answer

The inverse.
51554510
Consider the statement: “If \(\triangle ABC\) is isosceles with base \(\overline{AB}\), then the base angles \(\alpha\) and \(\beta\) are congruent.” a) Identify the hypothesis and conclusion. b) Write the converse. c) Determine whether the converse is true or false.

Hints

- The hypothesis follows “if,” and the conclusion follows “then.” - Form the converse by switching the hypothesis and conclusion. - Relate congruent angles in a triangle to their opposite sides.

Solution

1. The hypothesis is that \(\triangle ABC\) is isosceles with base \(\overline{AB}\). The conclusion is that \(\alpha = \beta\). 2. The converse is: “If \(\alpha = \beta\) in \(\triangle ABC\), then \(\triangle ABC\) is isosceles with base \(\overline{AB}\).” 3. The converse is true. In a triangle, congruent angles have congruent opposite sides. Therefore, \(AC = BC\), so \(\triangle ABC\) is isosceles with base \(\overline{AB}\).

Answer

a) Hypothesis: \(\triangle ABC\) is isosceles with base \(\overline{AB}\). Conclusion: \(\alpha = \beta\). b) If \(\alpha = \beta\), then \(\triangle ABC\) is isosceles with base \(\overline{AB}\). c) The converse is true.
51554610
Consider the statement: “If a quadrilateral is a square, then its diagonals are perpendicular.” a) Write the converse. b) Determine whether the converse is true. Justify your answer with an explanation or a counterexample.

Hints

- Switch the hypothesis and conclusion. - Look for another quadrilateral with perpendicular diagonals. - A counterexample must satisfy the new hypothesis but not its conclusion.

Solution

1. The converse is: “If a quadrilateral has perpendicular diagonals, then it is a square.” 2. The converse is false. 3. A rhombus that is not a square has perpendicular diagonals but does not have four right angles. Therefore, perpendicular diagonals alone do not guarantee that a quadrilateral is a square.

Answer

a) If a quadrilateral has perpendicular diagonals, then it is a square. b) The converse is false. A nonsquare rhombus is a counterexample.
54215610
Points \(A\) and \(B\) are distinct. Consider these statements about a point \(P\): I. If \(P\) lies on the perpendicular bisector of \(\overline{AB}\), then \(PA=PB\). II. If \(PA=PB\), then \(P\) lies on the perpendicular bisector of \(\overline{AB}\). III. If \(PA\ne PB\), then \(P\) does not lie on the perpendicular bisector of \(\overline{AB}\). a) Which statement is the converse of statement I? b) Which statement is the contrapositive of statement I? c) Which pair is logically equivalent, and which of the three statements are true in Euclidean geometry?
Figure for problem 542156

Hints

- Track what happens to the hypothesis and conclusion in each statement. - One logical form reverses the two parts; another also negates both parts. - Separate logical equivalence from whether a geometric converse happens to be true.

Solution

1. Statement II switches the hypothesis and conclusion of statement I, so II is the converse. 2. Statement III negates both parts of statement I and reverses their order, so III is the contrapositive. 3. A conditional statement and its contrapositive are logically equivalent, so statements I and III are equivalent. 4. The perpendicular-bisector theorem and its converse are both true: points on the perpendicular bisector are equidistant from the endpoints, and points equidistant from the endpoints lie on the perpendicular bisector. 5. Therefore, statements I, II, and III are all true.

Answer

a) Statement II. b) Statement III. c) Statements I and III are logically equivalent. All three statements are true.
54216410
Consider statement I: “If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.” Two related statements are: II. “If a quadrilateral is a parallelogram, then its diagonals bisect each other.” III. “If a quadrilateral is not a parallelogram, then its diagonals do not bisect each other.” a) Identify which statement is the converse of I and which is the contrapositive of I. b) State whether II and III are true. c) Which of II or III is logically equivalent to I? Explain.

Hints

- A converse reverses the two parts without negating them. - A contrapositive both reverses and negates the two parts. - Logical equivalence is automatic for one of these forms but not the other.

Solution

1. Statement II reverses the hypothesis and conclusion of I, so II is the converse. 2. Statement III negates and reverses the conclusion and hypothesis of I, so III is the contrapositive. 3. Statement II is true because diagonals of every parallelogram bisect each other. 4. Statement III is true because it is the contrapositive of the true statement I. 5. A conditional and its contrapositive are logically equivalent, so III is logically equivalent to I; the converse II is a separate statement even though it also happens to be true here.

Answer

a) II is the converse; III is the contrapositive. b) Both II and III are true. c) III is logically equivalent to I because a conditional and its contrapositive always have the same truth value.
54217110
Consider the statement: “If two coplanar lines cut by a transversal are parallel, then a pair of corresponding angles is congruent.” Farah writes, “If a pair of corresponding angles is congruent, then the two lines are parallel,” and labels it the contrapositive. a) Correct Farah’s label. b) Write the actual contrapositive of the original statement. c) State whether Farah’s statement and the actual contrapositive are true or false.
Figure for problem 542171

Hints

- Compare exactly how Farah changed the hypothesis and conclusion. - Check whether either part was negated. - Use the relevant parallel-line theorem and its converse to judge truth.

Solution

1. Farah’s statement switches the hypothesis and conclusion without negating them, so it is the converse. 2. The contrapositive reverses and negates both parts: If a pair of corresponding angles is not congruent, then the two lines are not parallel. 3. Farah’s converse is true by the converse of the corresponding angles theorem. 4. The actual contrapositive is true because it is logically equivalent to the true original statement.

Answer

a) Farah’s statement is the converse. b) If a pair of corresponding angles is not congruent, then the two lines are not parallel. c) Both the converse and the contrapositive are true.
54217810
A theorem states: “If a convex quadrilateral is cyclic, then each pair of opposite angles is supplementary.” A convex quadrilateral has one pair of opposite angles measuring \(103^\circ\) and \(75^\circ\). Three statements are proposed: A. If a pair of opposite angles is not supplementary, then the quadrilateral is not cyclic. B. If the quadrilateral is not cyclic, then a pair of opposite angles is not supplementary. C. If a pair of opposite angles is supplementary, then the quadrilateral is cyclic. a) Which proposed statement is the contrapositive of the theorem? b) Use the correct logical form to determine whether the given quadrilateral can be cyclic. c) Identify the logical form of statement B relative to the theorem.

Hints

- Represent the theorem as \(P\to Q\) before comparing the options. - Check the numerical angle condition before applying a logical implication. - The inverse negates both parts without reversing them.

Solution

1. The theorem has the form \(P\to Q\), where \(P\) is “the quadrilateral is cyclic” and \(Q\) is “each pair of opposite angles is supplementary.” 2. Its contrapositive is \(\neg Q\to\neg P\), which is statement A. 3. The two given opposite angles sum to \(103^\circ+75^\circ=178^\circ\), so they are not supplementary. 4. By statement A, the quadrilateral is not cyclic. 5. Statement B has form \(\neg P\to\neg Q\), so it is the inverse, not the contrapositive.

Answer

a) Statement A. b) The angles total \(178^\circ\), so they are not supplementary; therefore the quadrilateral is not cyclic. c) Statement B is the inverse.
54218210
A theorem states: “If a triangle is right with legs \(a,b\) and hypotenuse \(c\), then \(a^2+b^2=c^2\).” A triangle has side lengths \(8\), \(15\), and \(17\). a) Which logical form of the theorem must be true in order to conclude that this triangle is right: the original statement, converse, inverse, or contrapositive? b) Verify the needed hypothesis numerically and state the conclusion. c) In this theorem-and-converse pair, is the equation \(a^2+b^2=c^2\) a necessary condition, a sufficient condition, or both for a triangle to be right?

Hints

- Compare the information you are given with the hypothesis and conclusion of the stated theorem. - Use the longest side as the candidate hypotenuse. - Necessary and sufficient correspond to the two directions of a true biconditional.

Solution

1. The desired reasoning starts from the equation and concludes that the triangle is right, so it uses the converse. 2. The longest side is \(17\). Compute \(8^2+15^2=64+225=289=17^2\). 3. By the converse of the Pythagorean Theorem, the triangle is right. 4. The original theorem makes the equation necessary for a right triangle, and the true converse makes it sufficient. Therefore, within the stated side labeling, the equation is both necessary and sufficient.

Answer

a) The converse. b) \(8^2+15^2=17^2\), so the triangle is right. c) The equation is both necessary and sufficient for the triangle to be right.
54221210
A theorem says: “If a point lies inside an angle and on its angle bisector, then its perpendicular distances to the two sides are equal.” Point \(P\) lies inside an angle. Its perpendicular distances to the sides are \(4.2\,\text{cm}\) and \(5.1\,\text{cm}\). Maya concludes, “\(P\) is not on the angle bisector.” a) Is Maya's inference logically valid? b) Which logical form of the theorem justifies it? c) State the two facts in the problem that match the hypothesis of that logical form.

Hints

- Identify what would have to be equal if \(P\) were on the angle bisector. - Maya starts from a failure of the theorem's conclusion. - Ask which standard logical form starts from “not the conclusion.”

Solution

1. The theorem has form: if \(P\) is on the angle bisector, then the two perpendicular distances are equal. 2. Its contrapositive says: if the two perpendicular distances are not equal, then \(P\) is not on the angle bisector. 3. The distances \(4.2\,\text{cm}\) and \(5.1\,\text{cm}\) are unequal, and the point is inside the angle as required by the theorem's setting. 4. Therefore, Maya's conclusion is valid by the contrapositive.

Answer

a) Yes. b) The contrapositive. c) \(P\) lies inside the angle, and its two perpendicular distances are unequal: \(4.2\,\text{cm}\ne5.1\,\text{cm}\).
54222510
A theorem states: “In the same circle, if two chords are congruent, then their minor arcs are congruent.” Which statement is logically equivalent to the theorem? A. If two minor arcs are congruent, then their chords are congruent. B. If two chords are not congruent, then their minor arcs are not congruent. C. If two minor arcs are not congruent, then their chords are not congruent. Two minor arcs in one circle measure \(84^\circ\) and \(96^\circ\). Use the logically equivalent statement to compare their corresponding chords.

Hints

- Write the theorem mentally as \(P\to Q\). - Logical equivalence requires reversing and negating in a specific way. - The arc measures tell you directly whether \(Q\) is true or false.

Solution

1. Let \(P\) mean “the chords are congruent” and \(Q\) mean “the minor arcs are congruent.” 2. The contrapositive of \(P\to Q\) is \(\neg Q\to\neg P\), which is statement C. 3. A conditional and its contrapositive are logically equivalent. 4. Since \(84^\circ\ne96^\circ\), the minor arcs are not congruent. 5. By statement C, their corresponding chords are not congruent.

Answer

Statement C is logically equivalent to the theorem. Since the two minor arcs have different measures, their corresponding chords are not congruent.
54223310
The diagram shows chords \(\overline{AB}\) and \(\overline{CD}\) in the same circle with their perpendicular distances from center \(O\). A theorem says: “If two chords of the same circle are congruent, then they are equidistant from the center.” Federica says, “Because the distances are different, the chords are not congruent. I used the converse.” a) Is Federica’s geometric conclusion correct? b) Is Federica’s label “converse” correct? Identify the logical form actually used. c) The reverse theorem is also true. State what that additional fact lets you say about congruent chords and equal center distances as a biconditional.
Figure for problem 542233

Hints

- Compare “unequal distances” with the negation of the theorem’s conclusion. - Reversing a statement is different from reversing and negating it. - A biconditional needs the theorem and a true reverse direction.

Solution

1. Federica’s conclusion is correct: the center distances are \(6\,\text{cm}\) and \(9\,\text{cm}\), so they are unequal and the chords cannot be congruent. 2. The theorem has form \(P\to Q\), where \(P\) is congruent chords and \(Q\) is equal distances. The inference from unequal distances to noncongruent chords is \(\neg Q\to\neg P\), the contrapositive, not the converse. 3. The true reverse theorem says that chords equidistant from the center are congruent. 4. Together the two directions give: two chords of the same circle are congruent if and only if they are equidistant from the center.

Answer

a) Yes, the chords are not congruent. b) No. The inference uses the contrapositive. c) Two chords of the same circle are congruent if and only if they are equidistant from the center.
54224010
A theorem states: “If a parallelogram is a rhombus, then its diagonals are perpendicular.” A parallelogram has diagonals with slopes \(2\) and \(-\frac{1}{3}\). Elena concludes that the parallelogram is not a rhombus. a) Verify the geometric fact about the two diagonals that Elena is using. b) Is Elena's conclusion valid? c) Which logical form of the theorem justifies her conclusion: converse, inverse, or contrapositive?

Hints

- First determine whether the slopes satisfy the perpendicular-line criterion. - Compare the fact you obtain with the theorem's conclusion. - An inference beginning with the negation of a theorem's conclusion has one standard logical form.

Solution

1. Perpendicular nonvertical lines have slope product \(-1\). Here, \(2\cdot\left(-\frac13\right)=-\frac23\ne-1\), so the diagonals are not perpendicular. 2. The theorem has form: rhombus \(\to\) perpendicular diagonals. 3. Its contrapositive is: not perpendicular diagonals \(\to\) not a rhombus. 4. Therefore, Elena's conclusion is valid and uses the contrapositive.

Answer

a) The slope product is \(-\frac23\), so the diagonals are not perpendicular. b) Yes; the parallelogram is not a rhombus. c) The contrapositive.
54225410
Consider the statement: “If a quadrilateral is a parallelogram, then both pairs of opposite sides are congruent.” a) Write the converse and state whether it is true. b) Write the contrapositive. c) Quadrilateral \(WXYZ\) has \(WX=YZ=7\,\text{cm}\), \(XY=5\,\text{cm}\), and \(WZ=6\,\text{cm}\). What can you conclude from the contrapositive?

Hints

- Preserve the phrase “both pairs” when reversing the statement. - Negating “both pairs are congruent” means that at least one pair is not congruent. - Compare the lengths in each pair of opposite sides.

Solution

1. The converse is: if both pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram. This converse is true. 2. The contrapositive is: if at least one pair of opposite sides of a quadrilateral is not congruent, then the quadrilateral is not a parallelogram. 3. In \(WXYZ\), \(XY\ne WZ\) because \(5\ne6\). 4. Therefore, at least one pair of opposite sides is not congruent, so \(WXYZ\) is not a parallelogram.

Answer

a) If both pairs of opposite sides of a quadrilateral are congruent, then it is a parallelogram. The converse is true. b) If at least one pair of opposite sides of a quadrilateral is not congruent, then it is not a parallelogram. c) \(WXYZ\) is not a parallelogram.
54226110
In triangle \(ABC\), point \(M\) is the midpoint of side \(\overline{AB}\). Consider the statement: “If triangle \(ABC\) is right at \(C\), then \(MA=MB=MC\).” a) Write the converse. b) Write the contrapositive of the original statement. c) Suppose \(MA=5\,\text{cm}\) and \(MC=4.7\,\text{cm}\). What does the contrapositive show about \(\angle C\)?

Hints

- Treat the midpoint condition as fixed context while changing the conditional. - Negate the equality of all three distances carefully. - Compare the given distance from \(M\) to \(C\) with the half-hypotenuse distances.

Solution

1. The converse is: if \(MA=MB=MC\), then triangle \(ABC\) is right at \(C\). 2. The contrapositive is: if \(MA\), \(MB\), and \(MC\) are not all equal, then triangle \(ABC\) is not right at \(C\). 3. Since \(M\) is the midpoint of \(\overline{AB}\), \(MA=MB=5\,\text{cm}\), but \(MC=4.7\,\text{cm}\). The three distances are not all equal. 4. Therefore, \(\angle C\ne90^\circ\).

Answer

a) If the midpoint \(M\) of \(\overline{AB}\) is equidistant from \(A\), \(B\), and \(C\), then triangle \(ABC\) is right at \(C\). b) If those three distances from \(M\) are not all equal, then triangle \(ABC\) is not right at \(C\). c) \(\angle C\ne90^\circ\).
54226810
Points \(A\) and \(B\) lie on a circle, and point \(P\) lies outside the circle. Consider the statement: “If \(\overline{PA}\) and \(\overline{PB}\) are tangent segments, then \(PA=PB\).” a) Write the converse. b) Is the converse true? Give a counterexample. c) Write the contrapositive of the original statement.

Hints

- Reverse the hypothesis and conclusion while keeping the stated circle context fixed. - Look for two equal segments from an external point to points on a circle that are not tangent. - Reverse and negate both parts to form the contrapositive.

Solution

1. The converse is: if \(PA=PB\), then \(\overline{PA}\) and \(\overline{PB}\) are tangent segments. 2. The converse is false. For example, use the unit circle centered at \(O(0, 0)\), let \(P(2, 0)\), \(A(0, 1)\), and \(B(0, -1)\). Then \(P\) is outside the circle and \(PA=PB=\sqrt{5}\). 3. However, neither \(\overline{PA}\) nor \(\overline{PB}\) is perpendicular to the radius at its endpoint, so neither segment is tangent. 4. The contrapositive is: if \(PA\ne PB\), then \(\overline{PA}\) and \(\overline{PB}\) are not both tangent segments.

Answer

a) If \(PA=PB\), then \(\overline{PA}\) and \(\overline{PB}\) are tangent segments. b) False. On the unit circle, take \(P(2, 0)\), \(A(0, 1)\), and \(B(0, -1)\). Then \(PA=PB=\sqrt{5}\), but the segments are not tangent. c) If \(PA\ne PB\), then the two segments are not both tangent segments.
54227510
Consider the statement: “If a parallelogram has one right angle, then it is a rectangle.” a) Write the converse. b) Write the contrapositive. c) A parallelogram has an interior angle measuring \(110^\circ\). Use the contrapositive to determine whether it can have any right angle.

Hints

- Keep the condition that the figure is a parallelogram. - Reverse the original parts for the converse, and reverse-negate them for the contrapositive. - Use the angle relationships in a parallelogram to interpret the \(110^\circ\) angle.

Solution

1. The converse is: if a parallelogram is a rectangle, then it has at least one right angle. 2. The contrapositive is: if a parallelogram is not a rectangle, then it has no right angles. 3. A parallelogram with a \(110^\circ\) angle has adjacent angles of \(70^\circ\), so none of its angles is right. Equivalently, it is not a rectangle. 4. By the contrapositive, it cannot have any right angle.

Answer

a) If a parallelogram is a rectangle, then it has at least one right angle. b) If a parallelogram is not a rectangle, then it has no right angles. c) It cannot have any right angle.
54228210
Consider the statement: “If point \(P\) is the circumcenter of triangle \(ABC\), then \(PA=PB=PC\).” a) Write the converse and state whether it is true. b) Write the contrapositive. c) A point \(Q\) satisfies \(QA=7.2\,\text{cm}\), \(QB=7.2\,\text{cm}\), and \(QC=6.9\,\text{cm}\). What can you conclude?

Hints

- Reverse the defining condition for the converse. - Negate equality of all three distances carefully. - Compare the three given distances before applying the logical statement.

Solution

1. The converse is: if \(PA=PB=PC\), then \(P\) is the circumcenter of triangle \(ABC\). It is true because a point equidistant from all three vertices is the center of the circle through them. 2. The contrapositive is: if \(PA\), \(PB\), and \(PC\) are not all equal, then \(P\) is not the circumcenter of triangle \(ABC\). 3. For point \(Q\), the three distances are not all equal because \(6.9\ne7.2\). 4. Therefore, \(Q\) is not the circumcenter of triangle \(ABC\).

Answer

a) If \(PA=PB=PC\), then \(P\) is the circumcenter of triangle \(ABC\). The converse is true. b) If the three distances are not all equal, then \(P\) is not the circumcenter. c) \(Q\) is not the circumcenter of triangle \(ABC\).
54228910
Consider the statement: “If two nonvertical lines are perpendicular, then the product of their slopes is \(-1\).” a) Write the converse and state whether it is true. b) Write the contrapositive. c) Two nonvertical lines have slopes \(\frac{3}{4}\) and \(-\frac{4}{5}\). What can you conclude?
Figure for problem 542289

Hints

- Reverse the slope condition and geometric conclusion for the converse. - Reverse and negate both parts for the contrapositive. - Multiply the two given slopes exactly.

Solution

1. The converse is: if the product of the slopes of two nonvertical lines is \(-1\), then the lines are perpendicular. This converse is true. 2. The contrapositive is: if the product of the slopes of two nonvertical lines is not \(-1\), then the lines are not perpendicular. 3. The product of the given slopes is \(\frac{3}{4}\cdot\left(-\frac{4}{5}\right)=-\frac{3}{5}\). 4. Since \(-\frac{3}{5}\ne-1\), the lines are not perpendicular.

Answer

a) If the product of the slopes of two nonvertical lines is \(-1\), then the lines are perpendicular. The converse is true. b) If the product is not \(-1\), then the lines are not perpendicular. c) The lines are not perpendicular.
54229610
Consider the statement: “If parallelogram \(ABCD\) is a rhombus, then diagonal \(\overline{AC}\) bisects \(\angle A\) and \(\angle C\).” a) Write the converse and state whether it is true. b) Write the contrapositive. c) In a parallelogram, \(m\angle BAC=28^\circ\) and \(m\angle CAD=31^\circ\). What can you conclude?

Hints

- Keep the parallelogram condition in the converse. - Negate the statement that one diagonal bisects both named angles. - Compare the two parts of \(\angle A\).

Solution

1. The converse is: if diagonal \(\overline{AC}\) bisects \(\angle A\) and \(\angle C\) of a parallelogram, then the parallelogram is a rhombus. This converse is true. 2. The contrapositive is: if \(\overline{AC}\) fails to bisect at least one of \(\angle A\) and \(\angle C\), then the parallelogram is not a rhombus. 3. Since \(28^\circ\ne31^\circ\), diagonal \(\overline{AC}\) does not bisect \(\angle A\). 4. Therefore, the parallelogram is not a rhombus.

Answer

a) If \(\overline{AC}\) bisects \(\angle A\) and \(\angle C\) of a parallelogram, then it is a rhombus. The converse is true. b) If \(\overline{AC}\) does not bisect at least one of those angles, then the parallelogram is not a rhombus. c) The parallelogram is not a rhombus.
54231010
Consider the statement: “If three positive lengths form a triangle, then the sum of any two lengths is greater than the third length.” a) Write the converse and state whether it is true. b) Write a contrapositive form that is useful for testing three lengths. c) Can lengths \(3\,\text{cm}\), \(4\,\text{cm}\), and \(8\,\text{cm}\) form a triangle?

Hints

- Reverse the necessary condition to form the converse. - Negate “every pair has a sum greater than the third.” - Test the two smaller lengths against the largest length.

Solution

1. The converse is: if each pair of three positive lengths has a sum greater than the remaining length, then the three lengths form a triangle. This converse is true. 2. A useful contrapositive is: if the sum of some two lengths is less than or equal to the third length, then the three lengths do not form a triangle. 3. Here, \(3\,\text{cm}+4\,\text{cm}=7\,\text{cm}\le8\,\text{cm}\). 4. Therefore, the three lengths cannot form a triangle.

Answer

a) If all three triangle inequalities hold, then the lengths form a triangle. The converse is true. b) If one pair sums to no more than the third length, the lengths do not form a triangle. c) No, because \(3+4\le8\).
54231710
Consider the statement: “If two distinct circles are tangent at point \(T\), then their centers and \(T\) are collinear.” a) Write the converse and state whether it is true when the circles share point \(T\). b) Write the contrapositive. c) Two circles share point \(T(2, 1)\) and have centers \(O_1(0, 0)\) and \(O_2(5, 0)\). Can they be tangent at \(T\)?
Figure for problem 542317

Hints

- Relate each circle’s tangent at \(T\) to its radius through \(T\). - Reverse and negate the collinearity conclusion for the contrapositive. - Check whether the given point lies on the line through the centers.

Solution

1. The converse is: if two distinct circles share point \(T\) and their centers and \(T\) are collinear, then the circles are tangent at \(T\). This is true because both radii at \(T\) lie on the same line, so the circles have the same tangent line perpendicular to that line. 2. The contrapositive is: if the two centers and \(T\) are not collinear, then the circles are not tangent at \(T\). 3. Centers \(O_1\) and \(O_2\) lie on the x-axis, but \(T(2, 1)\) does not lie on the x-axis. 4. Therefore, the three points are not collinear, so the circles cannot be tangent at \(T\).

Answer

a) If two distinct circles share \(T\) and both centers are collinear with \(T\), then they are tangent at \(T\). The converse is true. b) If the centers and \(T\) are not collinear, the circles are not tangent at \(T\). c) No, the circles cannot be tangent at \(T\).
54232410
Consider the statement: “If quadrilateral \(ABCD\) is a kite with \(AB=AD\) and \(CB=CD\), then \(\angle B\cong\angle D\).” a) Write the converse. b) Is the converse true? Give a counterexample. c) Write the contrapositive of the original statement.
Figure for problem 542324

Hints

- Reverse the side-based hypothesis and angle conclusion for the converse. - Look for a familiar quadrilateral with congruent opposite angles but unequal adjacent sides. - Reverse and negate both parts for the contrapositive.

Solution

1. The converse is: if \(\angle B\cong\angle D\), then \(ABCD\) is a kite with \(AB=AD\) and \(CB=CD\). 2. The converse is false. A nonsquare rectangle has \(\angle B\cong\angle D\), but it does not have two pairs of congruent consecutive sides and is not a kite under the stated side condition. 3. The contrapositive is: if \(\angle B\not\cong\angle D\), then \(ABCD\) is not a kite with \(AB=AD\) and \(CB=CD\).

Answer

a) If \(\angle B\cong\angle D\), then \(ABCD\) is a kite with the stated side pairs. b) False; a nonsquare rectangle is a counterexample. c) If \(\angle B\not\cong\angle D\), then \(ABCD\) is not such a kite.
54233110
Consider the statement: “If two triangles are congruent, then all pairs of corresponding angles are congruent.” a) Write the converse. b) Is the converse true? Give a counterexample. c) Write the contrapositive of the original statement.

Hints

- Reverse the hypothesis and conclusion for the converse. - Look for triangles with the same shape but different sizes. - Negate the claim that every corresponding angle pair is congruent.

Solution

1. The converse is: if all pairs of corresponding angles of two triangles are congruent, then the triangles are congruent. 2. The converse is false. A \(3\)-\(4\)-\(5\) triangle and a \(6\)-\(8\)-\(10\) triangle have congruent corresponding angles because they are similar, but their corresponding side lengths differ, so they are not congruent. 3. The contrapositive is: if at least one pair of corresponding angles is not congruent, then the triangles are not congruent.

Answer

a) If all corresponding angles are congruent, then the triangles are congruent. b) False; \(3\)-\(4\)-\(5\) and \(6\)-\(8\)-\(10\) triangles are a counterexample. c) If at least one pair of corresponding angles is not congruent, then the triangles are not congruent.
54233810
In this problem, a trapezoid has exactly one pair of parallel sides. A theorem says: “If a trapezoid is isosceles, then its diagonals are congruent.” Its converse is also true. a) Combine the theorem and its converse into one biconditional statement. b) Write an equivalent biconditional using “not isosceles” and “not congruent.” c) A trapezoid has diagonals of lengths \(11\,\text{cm}\) and \(12\,\text{cm}\). What can you conclude?

Hints

- A biconditional is valid only when both a conditional and its converse are true. - For an equivalent negative form, negate both conditions in the biconditional. - Compare the two given diagonal lengths before applying the negative form.

Solution

1. Let \(P\) be “the trapezoid is isosceles” and let \(Q\) be “its diagonals are congruent.” The theorem is \(P\Rightarrow Q\), and its true converse is \(Q\Rightarrow P\). 2. Therefore, the two directions combine to give: a trapezoid is isosceles if and only if its diagonals are congruent. 3. Negating both equivalent conditions gives another biconditional: a trapezoid is not isosceles if and only if its diagonals are not congruent. 4. Since \(11\ne12\), the diagonals are not congruent. 5. Therefore, the trapezoid is not isosceles.

Answer

a) A trapezoid is isosceles if and only if its diagonals are congruent. b) A trapezoid is not isosceles if and only if its diagonals are not congruent. c) The trapezoid is not isosceles.
54235210
Consider the statement: “If point \(P\) lies outside a circle with center \(O\) and radius \(r\), then \(OP>r\).” a) Write the converse and state whether it is true. b) Write the contrapositive. c) A circle has radius \(8\,\text{cm}\), and \(OP=8.2\,\text{cm}\). Use one of the statements to classify \(P\).

Hints

- Reverse the condition and conclusion to form the converse. - Negate “outside” and the strict inequality carefully. - Compare the given center-to-point distance with the radius.

Solution

1. The converse is: if \(OP>r\), then \(P\) lies outside the circle. This is true by the distance definition of the exterior of a circle. 2. The contrapositive is: if \(OP\le r\), then \(P\) does not lie outside the circle. 3. Since \(8.2>8\), the converse applies, so \(P\) lies outside the circle.

Answer

a) If \(OP>r\), then \(P\) lies outside the circle. This is true. b) If \(OP\le r\), then \(P\) is not outside the circle. c) Point \(P\) lies outside the circle.
54235910
In a circle, consider the statement: “If a diameter is perpendicular to a chord, then it bisects the chord.” Assume the chord is not itself a diameter. a) Write the converse and state whether it is true. b) Write the contrapositive. c) Diameter \(\overline{CD}\) intersects chord \(\overline{AB}\) at \(M\), and \(AM=MB\). What can you conclude?

Hints

- Reverse the condition and conclusion without changing their meanings. - Negate “bisects” and “is perpendicular” carefully. - Interpret \(AM=MB\) as a statement about point \(M\).

Solution

1. The converse is: if a diameter bisects a chord that is not a diameter, then it is perpendicular to the chord. This is true. 2. The contrapositive is: if a diameter does not bisect a chord, then it is not perpendicular to the chord. 3. Since \(AM=MB\), diameter \(\overline{CD}\) bisects chord \(\overline{AB}\). By the converse, \(CD\perp AB\).

Answer

a) If a diameter bisects a chord that is not a diameter, then it is perpendicular to the chord. This is true. b) If a diameter does not bisect a chord, then it is not perpendicular to the chord. c) \(CD\perp AB\).
54237310
Consider the statement: “If quadrilateral \(ABCD\) is a parallelogram, then \(\angle A+\angle B=180^\circ\) and \(\angle B+\angle C=180^\circ\).” a) Write the converse and state whether it is true for a convex quadrilateral. b) Write the contrapositive. c) A convex quadrilateral has angle measures \(72^\circ\), \(108^\circ\), \(72^\circ\), and \(108^\circ\) in order. What can you conclude?

Hints

- Reverse the complete two-part conclusion, not just one angle condition. - Match each supplementary pair with a pair of opposite sides. - Check both required sums in the numerical case.

Solution

1. The converse is: if \(\angle A+\angle B=180^\circ\) and \(\angle B+\angle C=180^\circ\), then \(ABCD\) is a parallelogram. 2. The first supplementary pair implies \(AD\parallel BC\), and the second implies \(AB\parallel CD\). Thus the converse is true for a convex quadrilateral. 3. The contrapositive is: if \(ABCD\) is not a parallelogram, then \(\angle A+\angle B\ne180^\circ\) or \(\angle B+\angle C\ne180^\circ\). 4. For the given quadrilateral, \(72^\circ+108^\circ=180^\circ\) for both required consecutive pairs. By the converse, the quadrilateral is a parallelogram.

Answer

a) If both specified consecutive-angle pairs are supplementary, then the convex quadrilateral is a parallelogram. This is true. b) If the quadrilateral is not a parallelogram, then at least one specified pair is not supplementary. c) The quadrilateral is a parallelogram.
54238710
In this problem, a trapezoid has exactly one pair of parallel sides. Consider the statement: “If trapezoid \(ABCD\) with \(AB\parallel CD\) is isosceles, then \(\angle A\cong\angle B\).” a) Write the converse and state whether it is true. b) Write the contrapositive. c) In trapezoid \(ABCD\), \(\angle A=\angle B=68^\circ\). What can you conclude about the legs?

Hints

- Reverse the stated base-angle condition and isosceles conclusion. - Reverse and negate both parts for the contrapositive. - Apply the converse to the given base-angle equality.

Solution

1. The converse is: if \(\angle A\cong\angle B\) in trapezoid \(ABCD\) with \(AB\parallel CD\), then the trapezoid is isosceles. This is true. 2. The contrapositive is: if trapezoid \(ABCD\) is not isosceles, then \(\angle A\not\cong\angle B\). 3. Since \(\angle A=\angle B\), the converse applies. Therefore, \(ABCD\) is an isosceles trapezoid. 4. Its legs are congruent, so \(AD=BC\).

Answer

a) If \(\angle A\cong\angle B\) in trapezoid \(ABCD\) with \(AB\parallel CD\), then the trapezoid is isosceles. This is true. b) If the trapezoid is not isosceles, then \(\angle A\not\cong\angle B\). c) \(AD=BC\).
54239410
Consider the statement: “If a triangle is equilateral, then all three of its angles measure \(60^\circ\).” a) Write the converse and state whether it is true. b) Write the contrapositive. c) A triangle has angle measures \(59^\circ\), \(60^\circ\), and \(61^\circ\). What can you conclude about its side lengths?

Hints

- Reverse the side condition and the angle condition for the converse. - Negate “all three angles are \(60^\circ\)” carefully. - Compare unequal angles with their opposite sides.

Solution

1. The converse is: if all three angles of a triangle measure \(60^\circ\), then the triangle is equilateral. This is true because congruent angles have congruent opposite sides. 2. The contrapositive is: if at least one angle of a triangle does not measure \(60^\circ\), then the triangle is not equilateral. 3. The given triangle has angles that are not all \(60^\circ\), so the contrapositive shows it is not equilateral. 4. Since all three angle measures are different, the opposite side lengths are also all different. Thus the triangle is scalene.

Answer

a) If all three angles are \(60^\circ\), then the triangle is equilateral. This is true. b) If at least one angle is not \(60^\circ\), then the triangle is not equilateral. c) The triangle is scalene.
54240110
Consider the statement: “If triangle \(ABC\) is right at \(A\), then its orthocenter is \(A\).” a) Write the converse and state whether it is true. b) Write the contrapositive. c) The orthocenter of a triangle is vertex \(B\). What can you conclude?
Figure for problem 542401

Hints

- Reverse the right-angle condition and the orthocenter location. - Recall what it means for an altitude from another vertex to pass through \(A\). - Apply the same reasoning after relabeling the right-angle vertex.

Solution

1. The converse is: if the orthocenter of triangle \(ABC\) is \(A\), then the triangle is right at \(A\). 2. If the orthocenter is \(A\), the altitude from \(B\) passes through \(A\), so \(AB\perp AC\). Therefore, the converse is true. 3. The contrapositive is: if the orthocenter is not \(A\), then triangle \(ABC\) is not right at \(A\). 4. If the orthocenter is \(B\), applying the same converse with vertex \(B\) shows that \(BA\perp BC\). Thus the triangle is right at \(B\).

Answer

a) If the orthocenter is \(A\), then the triangle is right at \(A\). This is true. b) If the orthocenter is not \(A\), then the triangle is not right at \(A\). c) The triangle is right at \(B\).
54240810
In this problem, a trapezoid has exactly one pair of parallel sides. Trapezoid \(ABCD\) has \(\overline{AB}\parallel\overline{CD}\). Consider the statement: “If \(ABCD\) is cyclic, then \(AD=BC\).” a) Write the converse and state whether it is true. b) Write the contrapositive of the original statement. c) A trapezoid has leg lengths \(7\,\text{cm}\) and \(9\,\text{cm}\). What can you conclude about whether it is cyclic?

Hints

- Connect congruent legs in a trapezoid to its base angles. - For the contrapositive, reverse the statement and negate both parts. - Compare the two given leg lengths before choosing the useful logical form.

Solution

1. The converse is: If \(AD=BC\), then trapezoid \(ABCD\) is cyclic. This is true because congruent legs make the trapezoid isosceles, so its base angles are congruent and its opposite angles are supplementary. 2. The contrapositive is: If \(AD\ne BC\), then \(ABCD\) is not cyclic. 3. The given legs have different lengths, so the contrapositive applies.

Answer

a) If \(AD=BC\), then \(ABCD\) is cyclic. The converse is true. b) If \(AD\ne BC\), then \(ABCD\) is not cyclic. c) The trapezoid is not cyclic.
54241510
In triangle \(ABC\), point \(D\) lies on \(\overline{BC}\). Consider the statement: “If \(\overline{AD}\) bisects \(\angle A\), then \(\frac{BD}{DC}=\frac{AB}{AC}\).” a) Write the converse and state whether it is true. b) Write the contrapositive of the original statement. c) Suppose \(AB=8\,\text{cm}\), \(AC=12\,\text{cm}\), \(BD=6\,\text{cm}\), and \(DC=9\,\text{cm}\). What can you conclude about \(\overline{AD}\)?

Hints

- Reverse the hypothesis and conclusion without changing their mathematical meaning. - Negate an equality by writing an inequality. - Reduce both given ratios before applying a logical form.

Solution

1. The converse is: If \(\frac{BD}{DC}=\frac{AB}{AC}\), then \(\overline{AD}\) bisects \(\angle A\). This is the converse of the angle bisector theorem and is true. 2. The contrapositive is: If \(\frac{BD}{DC}\ne\frac{AB}{AC}\), then \(\overline{AD}\) does not bisect \(\angle A\). 3. The given ratios are \(\frac{BD}{DC}=\frac{6}{9}=\frac{2}{3}\) and \(\frac{AB}{AC}=\frac{8}{12}=\frac{2}{3}\). 4. Since the ratios are equal, the converse shows that \(\overline{AD}\) bisects \(\angle A\).

Answer

a) If \(\frac{BD}{DC}=\frac{AB}{AC}\), then \(\overline{AD}\) bisects \(\angle A\). The converse is true. b) If the two ratios are unequal, then \(\overline{AD}\) is not an angle bisector. c) Segment \(\overline{AD}\) bisects \(\angle A\).
54242210
Let \(A\) and \(B\) be endpoints of a diameter of a circle, and let \(P\) be distinct from \(A\) and \(B\). Consider the statement: “If \(P\) lies on the circle, then \(\angle APB=90^\circ\).” a) Write the converse and state whether it is true. b) Write the contrapositive. c) If \(m\angle APB=87^\circ\), what can you conclude about \(P\)?

Hints

- Reverse the original hypothesis and conclusion for the converse. - Negate “is a right angle” carefully. - Compare the measured angle with \(90^\circ\).

Solution

1. The converse is: If \(\angle APB=90^\circ\), then \(P\) lies on the circle with diameter \(\overline{AB}\). This is true by the converse of Thales' theorem. 2. The contrapositive is: If \(\angle APB\ne90^\circ\), then \(P\) does not lie on the circle. 3. Since \(87^\circ\ne90^\circ\), the contrapositive applies.

Answer

a) If \(\angle APB=90^\circ\), then \(P\) lies on the circle with diameter \(\overline{AB}\). The converse is true. b) If \(\angle APB\ne90^\circ\), then \(P\) is not on the circle. c) Point \(P\) is not on the circle.
54243510
Let \(c\) be the longest side of triangle \(ABC\), and let \(a\) and \(b\) be the other two sides. Consider the statement: “If triangle \(ABC\) is acute, then \(c^2<a^2+b^2\).” a) Write the converse and state whether it is true. b) Write the contrapositive. c) A triangle has side lengths \(5\), \(6\), and \(8\). What can you conclude?

Hints

- Identify the longest side before applying the inequality. - Negating a strict inequality changes its direction and includes equality. - Compare the square of the longest side with the sum of the other two squares.

Solution

1. The converse is: If \(c^2<a^2+b^2\), then triangle \(ABC\) is acute. This is true when \(c\) is the longest side. 2. The contrapositive is: If \(c^2\ge a^2+b^2\), then the triangle is not acute. 3. For the given triangle, the longest side is \(8\), and \(8^2=64\) while \(5^2+6^2=61\). 4. Since \(64>61\), the contrapositive shows that the triangle is not acute. The strict inequality rules out a right triangle, so the triangle is obtuse.

Answer

a) If \(c^2<a^2+b^2\), then the triangle is acute. The converse is true. b) If \(c^2\ge a^2+b^2\), then the triangle is not acute. c) The \(5\)-\(6\)-\(8\) triangle is obtuse.
54244910
A reflection across line \(m\) maps point \(P\) to point \(P''\). Consider the statement: “If \(P''=P\), then \(P\) lies on \(m\).” a) Write the converse and state whether it is true. b) Write the contrapositive of the original statement. c) Point \(Q\) is \(3\,\text{cm}\) from line \(m\). What can you conclude about its reflected image \(Q''\)?
Figure for problem 542449

Hints

- Identify the fixed points of a reflection. - Reverse the statement for the converse and negate both parts for the contrapositive. - Relate a point's distance from the reflection line to the distance from the point to its image.

Solution

1. The converse is: If \(P\) lies on \(m\), then reflection across \(m\) maps \(P\) to itself. This is true because every point on a reflection line is fixed. 2. The contrapositive is: If \(P\) does not lie on \(m\), then \(P''\ne P\). 3. Since \(Q\) is \(3\,\text{cm}\) from \(m\), it does not lie on \(m\), so \(Q''\ne Q\). 4. The reflection line is the perpendicular bisector of \(\overline{QQ''}\), so \(QQ''=2\cdot3=6\,\text{cm}\).

Answer

a) If \(P\) lies on \(m\), then \(P''=P\). The converse is true. b) If \(P\) is not on \(m\), then \(P''\ne P\). c) Point \(Q''\) is distinct from \(Q\), and \(QQ''=6\,\text{cm}\).
55596510
Rewrite the true statement “If a quadrilateral is a square, then it is a rectangle” using the words “sufficient” and “necessary.”

Hints

- “Sufficient” points from the hypothesis toward the conclusion. - “Necessary” describes a condition that must hold whenever the hypothesis holds. - Do not reverse the implication into “every rectangle is a square.”

Solution

1. In a true implication “If \(P\), then \(Q\),” condition \(P\) is sufficient for \(Q\). 2. Condition \(Q\) is necessary for \(P\). 3. Therefore, being a square is sufficient for being a rectangle, and being a rectangle is necessary for being a square.

Answer

Being a square is sufficient for being a rectangle. Being a rectangle is necessary for being a square.
55596610
Give coordinates for a convex quadrilateral whose diagonals are perpendicular but that is not a rhombus. Explain why your example disproves the converse of “If a quadrilateral is a rhombus, then its diagonals are perpendicular.”

Hints

- Make the diagonals easy to prove perpendicular, for example by placing them on the coordinate axes. - Then choose unequal distances from their intersection so the four side lengths are not all equal. - Verify both the required property and the failed conclusion.

Solution

1. One example is \(A(-2, 0)\), \(B(0, 3)\), \(C(2, 0)\), and \(D(0, -1)\), listed in order around the quadrilateral. 2. Diagonal \(\overline{AC}\) is horizontal and diagonal \(\overline{BD}\) is vertical, so the diagonals are perpendicular. 3. The squared side lengths are \(AB^2=13\), \(BC^2=13\), \(CD^2=5\), and \(DA^2=5\), so the quadrilateral is not a rhombus. 4. Thus perpendicular diagonals do not force a quadrilateral to be a rhombus, so the converse is false.

Answer

One valid example is \(A(-2, 0)\), \(B(0, 3)\), \(C(2, 0)\), \(D(0, -1)\). Its diagonals are perpendicular, but its side lengths are not all equal, so it is not a rhombus.
55596710
Jordan says the contrapositive of “If a quadrilateral is a square, then it is a rectangle” is “If a quadrilateral is a rectangle, then it is a square.” Identify Jordan's error and write the correct contrapositive.

Hints

- Compare Jordan's statement with the four logical forms \(P\to Q\), \(Q\to P\), \(\neg P\to\neg Q\), and \(\neg Q\to\neg P\). - A contrapositive requires both reversal and negation.

Solution

1. Jordan reversed the hypothesis and conclusion but did not negate either one, so Jordan wrote the converse. 2. The contrapositive must reverse and negate both parts. 3. The correct contrapositive is: “If a quadrilateral is not a rectangle, then it is not a square.”

Answer

Jordan wrote the converse, not the contrapositive. The correct contrapositive is: “If a quadrilateral is not a rectangle, then it is not a square.”
54218910
A circle has center \(O\) and radius \(6\,\text{cm}\). Line \(m\) passes through point \(T\), and \(OT\perp m\). Mekdes concludes that \(m\) is tangent to the circle at \(T\), even though it has not been established that \(T\) lies on the circle. a) Which logical form of the tangent-radius theorem is Mekdes trying to use? b) Explain why the conclusion is not yet justified, and state the missing condition. c) Write the contrapositive of the statement: “If \(m\) is tangent to the circle at \(T\), then \(OT\perp m\).”

Hints

- Compare the direction of the known theorem with Mekdes’s conclusion. - Check every condition required for a segment from the center to be a radius. - Form the contrapositive by reversing and negating both parts.

Solution

1. Mekdes is trying to use the converse: if a line is perpendicular to a radius at the radius’s endpoint on the circle, then the line is tangent there. 2. The given information does not show that \(T\) lies on the circle. The missing condition is \(OT=6\,\text{cm}\), equal to the circle’s radius. 3. With \(OT=6\,\text{cm}\), segment \(\overline{OT}\) is a radius ending at \(T\), and the perpendicular line \(m\) is tangent at \(T\). 4. The contrapositive is: If \(OT\) is not perpendicular to \(m\), then \(m\) is not tangent to the circle at \(T\).

Answer

a) The converse. b) It is not known that \(T\) lies on the circle. The missing condition is \(OT=6\,\text{cm}\). c) If \(OT\not\perp m\), then \(m\) is not tangent to the circle at \(T\).
54219610
A theorem states: “If a quadrilateral is a rectangle, then its diagonals are congruent.” Nneka claims that any quadrilateral with congruent diagonals must be a rectangle. a) Is Nneka’s claim logically the converse of the theorem? b) Show that the claim is false by giving a specific type of quadrilateral with congruent diagonals that need not be a rectangle. c) Add one geometric condition to “the diagonals are congruent” that makes a valid sufficient test for a rectangle, and explain why the added condition works.

Hints

- First decide whether Nneka reversed the theorem or negated it. - Look for a familiar quadrilateral class whose diagonals are congruent without forcing four right angles. - A valid repair should rule out that counterexample without simply saying “it is a rectangle.”

Solution

1. Yes. Nneka reverses the theorem’s hypothesis and conclusion, so the claim is the converse. 2. A nonrectangular isosceles trapezoid has congruent diagonals, so the converse is false for arbitrary quadrilaterals. 3. One valid repair is: “If a quadrilateral is a parallelogram and its diagonals are congruent, then it is a rectangle.” 4. A parallelogram with congruent diagonals satisfies a standard rectangle criterion, so the strengthened statement is true.

Answer

a) Yes; it is the converse. b) A nonrectangular isosceles trapezoid is a counterexample. c) For example, add the condition that the quadrilateral is a parallelogram. A parallelogram with congruent diagonals is a rectangle.
54220510
Consider the statement: “If two angles are vertical angles, then they are congruent.” Celine writes, “If two angles are not vertical angles, then they are not congruent,” and calls it the contrapositive. a) Correct Celine’s label and determine whether her statement is true. b) Write the actual contrapositive and state whether it is true. c) Write the converse and state whether it is true.
Figure for problem 542205

Hints

- Track whether the two parts were reversed, negated, or both. - Test false-looking statements with congruent angles from different locations. - Remember which logical form is always equivalent to the original conditional.

Solution

1. Celine’s statement negates both the hypothesis and conclusion without reversing them, so it is the inverse. 2. The inverse is false. Two nonvertical angles can still have the same measure. 3. The contrapositive is: If two angles are not congruent, then they are not vertical angles. 4. The contrapositive is true because it is logically equivalent to the original statement. 5. The converse is: If two angles are congruent, then they are vertical angles. 6. The converse is false because congruent angles can occur in many configurations other than a vertical pair.

Answer

a) It is the inverse, and it is false. b) If two angles are not congruent, then they are not vertical angles. This is true. c) If two angles are congruent, then they are vertical angles. This is false.
54221910
In triangle \(ABC\), point \(D\) is the midpoint of \(\overline{AB}\), and point \(E\) lies on \(\overline{AC}\). The statement “If \(DE\parallel BC\), then \(E\) is the midpoint of \(\overline{AC}\)” is true under these givens. a) Is the reverse implication also true? Justify geometrically. b) If both directions are true, write one biconditional statement combining them. c) In this setting, is \(DE\parallel BC\) necessary, sufficient, or both for \(E\) to be the midpoint of \(\overline{AC}\)?

Hints

- Keep the given fact that \(D\) is already the midpoint of \(AB\). - Recall what the segment joining two side midpoints does in a triangle. - A biconditional is justified only when both directions are true.

Solution

1. Yes. If \(D\) and \(E\) are the midpoints of two sides of a triangle, the Triangle Midsegment Theorem gives \(DE\parallel BC\). 2. Therefore, under the stated condition that \(D\) is the midpoint of \(AB\), the two statements are equivalent. 3. A valid biconditional is: “\(DE\parallel BC\) if and only if \(E\) is the midpoint of \(AC\).” 4. Because both directions are true, \(DE\parallel BC\) is both necessary and sufficient for \(E\) to be the midpoint in this setting.

Answer

a) Yes; the reverse is the Triangle Midsegment Theorem. b) Under the stated givens, \(DE\parallel BC\) if and only if \(E\) is the midpoint of \(\overline{AC}\). c) It is both necessary and sufficient.
54224710
In quadrilateral \(ABCD\), diagonal \(\overline{AC}\) is considered as a possible symmetry diagonal. Statement I: “If \(ABCD\) is a kite with \(AC\) as its symmetry diagonal, then \(AC\) is the perpendicular bisector of \(BD\).” Statement II: “If \(AC\) is the perpendicular bisector of \(BD\), then \(ABCD\) is a kite with \(AC\) as its symmetry diagonal.” a) Are both statements true? Justify statement II from the perpendicular-bisector property. b) If both are true, combine them into a biconditional. c) In the biconditional, is “\(AC\) is the perpendicular bisector of \(BD\)” necessary, sufficient, or both for the stated kite symmetry?

Hints

- Points on a perpendicular bisector are equidistant from the segment's endpoints. - Apply that locus fact separately to \(A\) and \(C\). - Once both directions are established, translate them into biconditional language.

Solution

1. Statement I is the standard symmetry property of a kite. 2. If \(AC\) is the perpendicular bisector of \(BD\), then points \(A\) and \(C\) lie on the perpendicular-bisector locus of \(BD\). Hence \(AB=AD\) and \(CB=CD\). 3. Reflection across \(AC\) swaps \(B\) and \(D\) while fixing \(A\) and \(C\), so the quadrilateral is a kite with \(AC\) as its symmetry diagonal. Thus statement II is also true. 4. Therefore: \(ABCD\) is a kite with symmetry diagonal \(AC\) if and only if \(AC\) is the perpendicular bisector of \(BD\). 5. Each condition is both necessary and sufficient for the other.

Answer

a) Yes. If \(AC\) perpendicularly bisects \(BD\), then \(AB=AD\) and \(CB=CD\), and reflection across \(AC\) swaps \(B\) and \(D\). b) \(ABCD\) is a kite with symmetry diagonal \(AC\) if and only if \(AC\) is the perpendicular bisector of \(BD\). c) It is both necessary and sufficient.
54230310
Consider the statement: “If a parallelogram is a square, then its diagonals are both congruent and perpendicular.” a) Write the converse and state whether it is true. b) Write the contrapositive of the original statement. c) A parallelogram has congruent diagonals, but their slopes are \(2\) and \(-1\). What can you conclude by using the contrapositive?

Hints

- Preserve both diagonal properties when reversing the statement. - Negate a conclusion joined by “and” by using “or.” - Test the given slopes for perpendicularity before applying the contrapositive.

Solution

1. The converse is: if a parallelogram has diagonals that are both congruent and perpendicular, then it is a square. This converse is true. 2. The contrapositive of the original statement is: if a parallelogram's diagonals are not congruent or are not perpendicular, then the parallelogram is not a square. 3. The product of the given diagonal slopes is \(2\cdot(-1)=-2\), not \(-1\), so the diagonals are not perpendicular. 4. Therefore, one of the required square properties fails. By the contrapositive, the parallelogram is not a square.

Answer

a) If a parallelogram has diagonals that are both congruent and perpendicular, then it is a square. The converse is true. b) If a parallelogram's diagonals are not congruent or are not perpendicular, then it is not a square. c) The slope product is \(-2\), so the diagonals are not perpendicular. Therefore, the parallelogram is not a square.
54234510
Use proof by contrapositive to prove this statement: “If a triangle’s circumcenter lies inside the triangle, then the triangle is acute.” Then classify a triangle whose circumcenter lies on one of its sides.

Hints

- For a contrapositive proof, negate the conclusion first and prove the negation of the hypothesis. - Split “not acute” into the right-triangle and obtuse-triangle cases. - Use the standard circumcenter location for each triangle type.

Solution

1. Let \(P\) be “the circumcenter lies inside the triangle” and let \(Q\) be “the triangle is acute.” The contrapositive of \(P\Rightarrow Q\) is: if the triangle is not acute, then its circumcenter does not lie inside the triangle. 2. A nonacute triangle is either right or obtuse. 3. In a right triangle, the circumcenter is the midpoint of the hypotenuse, so it lies on a side rather than inside the triangle. 4. In an obtuse triangle, the circumcenter lies outside the triangle. 5. Thus, whenever the triangle is not acute, its circumcenter is not inside. The contrapositive is true, so the original statement is true. 6. If the circumcenter lies on one of the triangle’s sides, the triangle is right.

Answer

The contrapositive is: if a triangle is not acute, then its circumcenter does not lie inside the triangle. A right triangle’s circumcenter lies on its hypotenuse, and an obtuse triangle’s circumcenter lies outside, so the contrapositive—and therefore the original statement—is true. A circumcenter on a side identifies a right triangle.
54236610
Use the hypothesis “two distinct circles are internally tangent” to write a true conditional whose conclusion describes the number of common points. a) Write the conditional. b) Write its converse and show that the converse is false with a geometric counterexample. c) Write the contrapositive. d) Two circles have radii \(9\,\text{cm}\) and \(4\,\text{cm}\), and their centers are \(13\,\text{cm}\) apart. Explain how this example relates to the converse.

Hints

- Recall how many common points tangent circles have. - Test the converse against both kinds of tangency, not only internal tangency. - For the numerical example, compare the center distance with the sum of the radii.

Solution

1. A true conditional is: if two distinct circles are internally tangent, then they have exactly one common point. 2. Its converse is: if two distinct circles have exactly one common point, then they are internally tangent. 3. The converse is false because externally tangent circles also have exactly one common point. 4. The contrapositive of the original conditional is: if two distinct circles do not have exactly one common point, then they are not internally tangent. 5. For the given circles, \(9+4=13\), which equals the center distance. Therefore, the circles are externally tangent. 6. They have exactly one common point but are not internally tangent, so they are a counterexample to the converse.

Answer

a) If two distinct circles are internally tangent, then they have exactly one common point. b) Converse: If two distinct circles have exactly one common point, then they are internally tangent. This is false; externally tangent circles are a counterexample. c) If two distinct circles do not have exactly one common point, then they are not internally tangent. d) Since \(13=9+4\), the circles are externally tangent. They have one common point but are not internally tangent, so they disprove the converse.
54238010
For a convex quadrilateral with diagonal lengths \(d_1\) and \(d_2\), consider the statement: “If the diagonals are perpendicular, then the area is \(\frac{1}{2}d_1d_2\).” a) Write the converse and state whether it is true. b) Write the contrapositive. c) A convex quadrilateral has diagonals of lengths \(12\,\text{cm}\) and \(9\,\text{cm}\) and area \(54\,\text{cm}^2\). What can you conclude?

Hints

- Reverse the perpendicularity and area statements for the converse. - Compare the special area formula with the formula involving the angle between diagonals. - Evaluate one-half the product of the given diagonal lengths.

Solution

1. The converse is: if the area of a convex quadrilateral is \(\frac{1}{2}d_1d_2\), then its diagonals are perpendicular. 2. In general, \(A=\frac{1}{2}d_1d_2\sin\theta\), where \(\theta\) is the angle between the diagonals. Equality with \(\frac{1}{2}d_1d_2\) requires \(\sin\theta=1\), so \(\theta=90^\circ\). Thus the converse is true. 3. The contrapositive is: if the area is not \(\frac{1}{2}d_1d_2\), then the diagonals are not perpendicular. 4. Here, \(\frac{1}{2}\cdot12\cdot9=54\,\text{cm}^2\), which equals the given area. By the converse, the diagonals are perpendicular.

Answer

a) If the area is \(\frac{1}{2}d_1d_2\), then the diagonals are perpendicular. This is true for a convex quadrilateral. b) If the area is not \(\frac{1}{2}d_1d_2\), then the diagonals are not perpendicular. c) The diagonals are perpendicular.
54242910
From point \(P\), one ray contains points \(A\) and \(B\), and another ray contains points \(C\) and \(D\), with the nearer point named first on each ray. Consider the statement: “If \(A\), \(B\), \(C\), and \(D\) lie on one circle, then \(PA\cdot PB=PC\cdot PD\).” a) Write the converse and state whether it is true. b) Write the contrapositive. c) Suppose \(PA=3\), \(PB=8\), \(PC=4\), and \(PD=6\). What can you conclude?
Figure for problem 542429

Hints

- Reverse the original theorem without changing the order of points on the rays. - For the contrapositive, negate the product equality first, then negate concyclicity. - Compute the two products before selecting the useful logical form.

Solution

1. The converse is: If \(PA\cdot PB=PC\cdot PD\), then \(A\), \(B\), \(C\), and \(D\) lie on one circle. With the stated ray order and distinct points, this converse of the secant-product theorem is true. 2. The contrapositive is: If \(PA\cdot PB\ne PC\cdot PD\), then the four points are not concyclic. 3. The products are \(3\cdot8=24\) and \(4\cdot6=24\). 4. Since the products are equal, the converse shows that the four points are concyclic.

Answer

a) Equal secant products imply that the four points are concyclic; the converse is true under the stated conditions. b) If the products are unequal, then the four points are not concyclic. c) The four points lie on one circle.
54244210
Consider a convex quadrilateral with four distinct vertices. Use the statement: “If the quadrilateral is a square, then a \(90^\circ\) rotation about the intersection of its diagonals maps the quadrilateral onto itself.” a) Write the converse and state whether it is true. b) Write the contrapositive. c) A quadrilateral is not mapped onto itself by a \(90^\circ\) rotation about any point. What can you conclude?

Hints

- Track where one vertex goes under repeated quarter-turns. - Compare all four distances from the rotation center and all four central angles. - Use the contrapositive for the final conclusion.

Solution

1. The converse is: If a \(90^\circ\) rotation about the intersection of a convex quadrilateral's diagonals maps the quadrilateral onto itself, then the quadrilateral is a square. 2. The converse is true. The rotation cycles the four vertices, so they are equally distant from the rotation center and consecutive central angles are \(90^\circ\). Thus, the vertices form a square. 3. The contrapositive is: If the quadrilateral is not mapped onto itself by that \(90^\circ\) rotation, then it is not a square. 4. The given quadrilateral fails the rotational condition about every point, so in particular it fails at any possible diagonal intersection. Therefore, it is not a square.

Answer

a) A convex quadrilateral invariant under a \(90^\circ\) rotation about its diagonal intersection is a square. The converse is true. b) If that rotation does not map the quadrilateral onto itself, then the quadrilateral is not a square. c) The quadrilateral is not a square.

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