Free math worksheets

- Separate the direction requirement from the point-of-passage requirement. - Compare how each of the two relevant lines relates to the same third line. - Check the point named in the final construction step.

1. Line \(m\) is perpendicular to \(\ell\), and line \(n\) is also perpendicular to \(m\). 2. Two coplanar lines perpendicular to the same line are parallel, so \(n\parallel\ell\). 3. Line \(n\) passes through \(Q\), not through \(P\), so it does not satisfy the requirement that the parallel line pass through \(P\). 4. In step 3, the perpendicular to \(m\) must be created through \(P\) rather than through \(Q\).

a) Yes. Both \(n\) and \(\ell\) are perpendicular to \(m\), so \(n\parallel\ell\). b) No. It is parallel to \(\ell\), but it does not pass through \(P\). c) Create the second perpendicular through \(P\), not through \(Q\).

- Translate each midpoint or reflection statement into a diagonal relationship. - Identify the quadrilateral whose diagonals meet at \(M\). - Use its classification to obtain the required pair of parallel sides.

1. Since \(M\) is the midpoint of \(\overline{BP}\), \(BM=MP\). 2. Since \(Q\) is the reflection of \(A\) across \(M\), \(M\) is also the midpoint of \(\overline{AQ}\). 3. In quadrilateral \(ABQP\), diagonals \(\overline{AQ}\) and \(\overline{BP}\) bisect each other at \(M\). 4. Therefore, \(ABQP\) is a parallelogram. 5. Opposite sides of a parallelogram are parallel, so \(PQ\parallel AB\). 6. Since \(\overline{AB}\) lies on \(\ell\), \(PQ\parallel\ell\).

The construction makes \(M\) the midpoint of both diagonals \(\overline{AQ}\) and \(\overline{BP}\) of quadrilateral \(ABQP\). The quadrilateral is a parallelogram, so \(PQ\parallel AB\), and therefore \(PQ\parallel\ell\).

- Identify how the two given angles are positioned relative to the transversal. - Compare their sum with the measure of a straight angle. - Recall a converse theorem that can establish that two lines are parallel.

1. The two same-side interior angles have sum \(68^\circ+112^\circ=180^\circ\). 2. If same-side interior angles formed by a transversal are supplementary, then the two lines are parallel. 3. Therefore, \(n\parallel\ell\).

Since \(68^\circ+112^\circ=180^\circ\), the same-side interior angles are supplementary. By the converse of the same-side interior angles theorem, \(n\parallel\ell\).

- Relate each constructed line to the original line \(\ell\). - Use a theorem about two lines perpendicular to the same line. - Locate the midpoints of the two adjacent equal segments to determine the separation.

1. Since \(\overline{AB}\) and \(\overline{BC}\) lie on \(\ell\), both perpendicular bisectors are perpendicular to \(\ell\). 2. Two distinct lines perpendicular to the same line are parallel, so \(p\parallel q\). 3. The midpoint of \(\overline{AB}\) is \(2.4\,\text{cm}\) from \(A\), and the midpoint of \(\overline{BC}\) is \(2.4\,\text{cm}\) from \(B\). 4. The distance between these midpoints is \(2.4\,\text{cm}+2.4\,\text{cm}=4.8\,\text{cm}\). This is the perpendicular distance between \(p\) and \(q\).

a) \(p\parallel q\) because both are perpendicular to \(\ell\). b) The distance between \(p\) and \(q\) is \(4.8\,\text{cm}\).

- Recall the relationship between a chord and the segment joining its midpoint to the center. - Compare that relationship with the perpendicular created at \(O\). - Use a theorem about two lines perpendicular to the same line.

1. Since \(M\) is the midpoint of chord \(\overline{AB}\), the segment from the center to \(M\) is perpendicular to the chord. Thus \(OM\perp AB\). 2. The app constructs \(n\perp OM\). 3. Two distinct lines perpendicular to the same line are parallel. 4. Therefore, \(n\parallel AB\).

The center-to-midpoint segment \(\overline{OM}\) is perpendicular to chord \(\overline{AB}\), and line \(n\) is also perpendicular to \(\overline{OM}\). Therefore, \(n\parallel AB\).

- Use the two equal angles created by the angle bisector. - Transfer one of those angle measures using the parallel through \(D\). - Identify the resulting isosceles triangle.

1. Since \(\overline{AD}\) bisects \(\angle A\), \(\angle BAD\cong\angle DAC\). 2. Because \(\overline{DE}\parallel\overline{AC}\), \(\angle ADE\cong\angle DAC\) by alternate interior angles. 3. Since \(E\) lies on \(\overline{AB}\), \(\angle EAD=\angle BAD\). 4. Therefore, \(\angle EAD\cong\angle ADE\). 5. In triangle \(AED\), congruent angles have congruent opposite sides, so \(AE=DE\).

The angle-bisector condition and the app-created parallel give \(\angle EAD=\angle ADE\). Therefore, triangle \(AED\) is isosceles and \(AE=DE\).

- Identify which side must be parallel to \(AC\). - Identify which side must be parallel to \(AB\). - Use the uniqueness of a parallel through an external point.

1. The app creates the line through \(B\) parallel to \(\overline{AC}\). 2. The app creates the line through \(C\) parallel to \(\overline{AB}\). 3. Let these two lines intersect at \(D\). 4. Then \(BD\parallel AC\) and \(CD\parallel AB\), so both pairs of opposite sides of quadrilateral \(ABDC\) are parallel. Therefore, \(ABDC\) is a parallelogram. 5. Through a point not on a given line, exactly one parallel to that line can be drawn. Thus both lines are unique, and their intersection \(D\) is unique.

The app creates the line through \(B\) parallel to \(AC\) and the line through \(C\) parallel to \(AB\). Their intersection is the unique point \(D\). The two pairs of opposite sides are parallel, so \(ABDC\) is a parallelogram.

- Decide which line through \(P\) must be parallel to \(OA\). - Decide which line through \(P\) must be parallel to \(OB\). - Use the uniqueness of a parallel through an external point.

1. Through \(P\), the app creates the line parallel to \(m\). Let it intersect \(\ell\) at \(A\). 2. Through \(P\), the app creates the line parallel to \(\ell\). Let it intersect \(m\) at \(B\). 3. Since \(OA\) lies on \(\ell\), \(OA\parallel PB\). Since \(OB\) lies on \(m\), \(OB\parallel AP\). 4. Both pairs of opposite sides of quadrilateral \(OAPB\) are parallel, so \(OAPB\) is a parallelogram. 5. Through \(P\), there is exactly one line parallel to \(m\) and exactly one line parallel to \(\ell\). Their intersections with \(\ell\) and \(m\) therefore determine unique points \(A\) and \(B\).

The app creates a parallel to \(m\) through \(P\) to locate \(A\) on \(\ell\), and a parallel to \(\ell\) through \(P\) to locate \(B\) on \(m\). Then \(OA\parallel PB\) and \(OB\parallel AP\), so \(OAPB\) is the unique required parallelogram.

- Identify the transversal and compare the locations of the two equal angles. - Decide which angle-pair relationship supports a converse theorem about parallel lines. - Check how changing sides of the transversal changes the geometric configuration.

1. The original angle at \(A\) and the copied angle at \(P\) lie between lines \(\ell\) and \(n\) on opposite sides of transversal \(\overline{AP}\). 2. The angles are congruent by the copying procedure, so they form a pair of congruent alternate interior angles. 3. By the converse of the alternate interior angles theorem, \(n\parallel\ell\). 4. If the angle is copied on the same side of the transversal as the original angle, the two congruent angles are not alternate interior angles. No parallel-line converse applies, and the resulting line generally intersects \(\ell\).

a) The copied angle and the original angle are congruent alternate interior angles, so the converse theorem gives \(n\parallel\ell\). b) The same-side placement does not create congruent alternate interior angles. It therefore does not justify a parallel line and generally produces a line that intersects \(\ell\).

- Compare the two segments that measure the offset from the original line. - Look for a quadrilateral test involving one pair of opposite sides. - Once the quadrilateral is classified, use its remaining opposite sides.

1. Since \(PF\perp\ell\) and \(GH\perp\ell\), segments \(\overline{PF}\) and \(\overline{GH}\) are parallel. 2. The transferred length gives \(PF=GH\). 3. In quadrilateral \(FGHP\), one pair of opposite sides, \(\overline{PF}\) and \(\overline{GH}\), is both parallel and congruent. 4. Therefore, \(FGHP\) is a parallelogram. 5. Opposite sides of a parallelogram are parallel, so \(PH\parallel FG\). Because \(\overline{FG}\) lies on \(\ell\), \(PH\parallel\ell\).

The equal perpendicular offsets make \(\overline{PF}\) and \(\overline{GH}\) parallel and congruent. Thus \(FGHP\) is a parallelogram, so \(PH\parallel FG\), and therefore \(PH\parallel\ell\).

- Create equal reference intervals on a ray from one endpoint. - Connect the last reference point to the other endpoint of the segment. - Use parallel lines to transfer the equal spacing to \(\overline{AB}\).

1. The app creates an auxiliary ray from \(A\) that is not collinear with \(\overline{AB}\). 2. Using one fixed length, the app marks five consecutive congruent segments on the ray, ending at points \(P_1,P_2,P_3,P_4,P_5\). 3. The app draws \(\overline{P_5B}\). 4. Through each of \(P_1,P_2,P_3,P_4\), the app creates a line parallel to \(\overline{P_5B}\). Let these lines meet \(\overline{AB}\) at \(Q_1,Q_2,Q_3,Q_4\). 5. Parallel lines cut transversals proportionally. Since the five segments on the auxiliary ray are congruent, the corresponding segments \(AQ_1\), \(Q_1Q_2\), \(Q_2Q_3\), \(Q_3Q_4\), and \(Q_4B\) are congruent.

The app marks five equal steps on an auxiliary ray from \(A\), connects the fifth point to \(B\), and creates parallels through the first four marked points. Their intersections divide \(\overline{AB}\) into five congruent parts by the proportional-segments theorem.

- Relate the copied chord equality to intercepted arcs. - Identify two inscribed angles that use those arcs. - View \(\overline{AC}\) as a transversal of the two chord lines.

1. Because the app marks \(AD=BC\), the congruent chords intercept congruent arcs. Thus arc \(AD\) is congruent to arc \(BC\). 2. Inscribed angle \(\angle ACD\) intercepts arc \(AD\), and inscribed angle \(\angle CAB\) intercepts arc \(CB\). 3. Congruent arcs have congruent inscribed angles, so \(\angle ACD\cong\angle CAB\). 4. These angles are alternate interior angles formed by transversal \(\overline{AC}\) with lines \(CD\) and \(AB\). 5. By the converse of the alternate interior angles theorem, \(CD\parallel AB\).

Since \(AD=BC\), arcs \(AD\) and \(BC\) are congruent, so \(\angle ACD=\angle CAB\). These are alternate interior angles, hence \(CD\parallel AB\).

- Compare the area formulas for a triangle and a rectangle with the same base. - Express the rectangle's height in terms of the triangle's altitude. - Use the midpoint condition on \(\overline{AH}\).

1. Let \(AH=h\). Since \(M\) is the midpoint of \(\overline{AH}\), \(MH=\frac{h}{2}\). 2. Because \(EF\parallel BC\) and \(BE\perp BC\), the height of rectangle \(BEFC\) is \(BE=MH=\frac{h}{2}\). 3. Therefore, \([BEFC]=BC\cdot\frac{h}{2}=\frac{1}{2}BC\cdot h\). 4. The area of triangle \(ABC\) is \([ABC]=\frac{1}{2}BC\cdot h\). 5. Hence \([BEFC]=[ABC]\).

The rectangle has the same base \(BC\) as the triangle and height \(\frac{1}{2}AH\). Therefore, both areas equal \(\frac{1}{2}BC\cdot AH\).

- The altitude condition determines a line parallel to the base. - A neighboring vertex of the rhombus must also be one side length from \(A\). - Complete each possibility using the second pair of parallel sides.

1. On the specified side of \(AB\), the app creates a line \(g\) parallel to \(AB\) at perpendicular distance \(6\,\text{cm}\). 2. The app creates the circle centered at \(A\) with radius \(10\,\text{cm}\). Because the distance from \(A\) to \(g\) is \(6\,\text{cm}<10\,\text{cm}\), the circle meets \(g\) at two points, \(D_1\) and \(D_2\). 3. For each \(D_i\), the app creates through \(B\) a line parallel to \(AD_i\). Let it meet \(g\) at \(C_i\). 4. Since \(AB\parallel C_iD_i\) and \(AD_i\parallel BC_i\), quadrilateral \(ABC_iD_i\) is a parallelogram. 5. Also, \(AD_i=AB=10\,\text{cm}\). A parallelogram with congruent consecutive sides is a rhombus. 6. The two intersections \(D_1\) and \(D_2\) produce two distinct rhombi, and no other point on \(g\) is \(10\,\text{cm}\) from \(A\).

The app creates the parallel line \(6\,\text{cm}\) from \(AB\). Its two intersections with the circle centered at \(A\) of radius \(10\,\text{cm}\) determine the two possible vertices \(D\); a parallel through \(B\) completes each rhombus. These are the only two solutions.

- Use the two parallel pairs to classify the inner quadrilateral. - Compare \(\triangle BDE\) with the original triangle. - Use the midpoint to determine the similarity scale factor.

1. Since \(AD\) lies on \(AB\) and \(EF\parallel AB\), \(AD\parallel EF\). Since \(AF\) lies on \(AC\) and \(DE\parallel AC\), \(AF\parallel DE\). Thus \(ADEF\) is a parallelogram. 2. Because \(AB\perp AC\), adjacent sides \(AD\) and \(AF\) are perpendicular. Therefore, \(ADEF\) is a rectangle. 3. Point \(D\) is the midpoint of \(AB\), so \(AD=DB=5\,\text{cm}\). 4. In \(\triangle BDE\), \(DE\parallel AC\), so \(\triangle BDE\sim\triangle BAC\). The scale factor is \(\frac{BD}{BA}=\frac{1}{2}\), giving \(DE=\frac{1}{2}AC=5\,\text{cm}\). 5. The rectangle has adjacent sides \(AD=DE=5\,\text{cm}\), so it is a square. Its area is \(5^2=25\,\text{cm}^2\).

The app-created parallels make \(ADEF\) a rectangle. Since \(AD=DE=5\,\text{cm}\), it is a square with area \(25\,\text{cm}^2\).

- Convert the area requirement into an altitude requirement. - Think of all points at one fixed perpendicular distance from a line. - After choosing one adjacent vertex, use parallel lines to complete the parallelogram.

1. For a parallelogram with base \(8\,\text{cm}\) and area \(40\,\text{cm}^2\), the required altitude is \(40\div8=5\,\text{cm}\). 2. The app creates a perpendicular to \(AB\) at \(A\) and marks point \(E\) on the chosen side so that \(AE=5\,\text{cm}\). 3. Through \(E\), the app creates line \(m\parallel AB\). 4. Choose any point \(D\) on \(m\). The app uses parallels through \(D\) and \(B\) to complete parallelogram \(ABCD\). 5. The altitude to base \(AB\) is \(5\,\text{cm}\), so the area is \(8\cdot5=40\,\text{cm}^2\). 6. Conversely, every parallelogram with the required area must have altitude \(5\,\text{cm}\), so its opposite base and its fourth vertex \(D\) lie on line \(m\).

The app creates the line parallel to \(AB\) at perpendicular distance \(5\,\text{cm}\). Every point on that line can serve as the fourth vertex \(D\), and every valid parallelogram has its opposite base on that same line.

- Place two of the given lengths on one ray and the third on another ray. - A parallel through the farther point creates a pair of similar triangles. - Match the sides in the similarity proportion to the requested ratio.

1. The app creates two rays from point \(O\). On the first ray, it marks \(A\) and \(B\) so that \(OA=3\,\text{cm}\) and \(OB=5\,\text{cm}\). 2. On the second ray, the app marks \(C\) so that \(OC=7\,\text{cm}\). 3. The app draws \(\overline{AC}\) and creates the line through \(B\) parallel to \(AC\), meeting the second ray at \(D\). 4. Triangles \(OAC\) and \(OBD\) are similar, so \(\frac{OA}{OB}=\frac{OC}{OD}\). 5. Therefore, \(\frac{3}{5}=\frac{7}{OD}\), giving \(OD=\frac{35}{3}\,\text{cm}\).

The app-created fourth proportional is \(x=\frac{35}{3}\,\text{cm}\).

- Use each side of the midpoint triangle to determine the direction of an unknown side. - After creating the three side lines, identify parallelograms that contain \(D\), \(E\), and \(F\). - Use opposite sides of those parallelograms to prove the midpoint equalities and then justify uniqueness.

1. In any triangle, the segment joining two side midpoints is parallel to the third side. 2. Through \(D\), the app creates the line parallel to \(EF\); this is line \(AB\). 3. Through \(E\), the app creates the line parallel to \(DF\); this is line \(BC\). 4. Through \(F\), the app creates the line parallel to \(DE\); this is line \(CA\). 5. Let \(A\) be the intersection of the first and third lines, \(B\) the intersection of the first and second, and \(C\) the intersection of the second and third. 6. Quadrilateral \(ADEF\) is a parallelogram because \(AD\parallel EF\) and \(AF\parallel DE\). Therefore, \(AD=EF\) and \(AF=DE\). 7. Quadrilateral \(DBEF\) is a parallelogram because \(DB\parallel EF\) and \(BE\parallel DF\). Therefore, \(DB=EF\) and \(BE=DF\). 8. Quadrilateral \(DECF\) is a parallelogram because \(DE\parallel CF\) and \(DF\parallel EC\). Therefore, \(EC=DF\) and \(CF=DE\). 9. Thus, \(AD=DB\), \(BE=EC\), and \(CF=FA\), so \(D\), \(E\), and \(F\) are the required side midpoints. 10. Each side line is uniquely determined by one given point and one required parallel direction, so their pairwise intersections and the reconstructed triangle are unique.

The app creates the three side lines through \(D\), \(E\), and \(F\), parallel respectively to \(EF\), \(DF\), and \(DE\). Their pairwise intersections are \(A\), \(B\), and \(C\). The parallelograms \(ADEF\), \(DBEF\), and \(DECF\) give \(AD=DB\), \(BE=EC\), and \(CF=FA\), so the given points are the side midpoints. The three uniquely determined side lines make the reconstruction unique.

- Split the quadrilateral along a diagonal. - Compare two triangles that share that diagonal as a base. - Vertices on a line parallel to a base give equal altitudes.

1. The app draws diagonal \(\overline{AC}\). 2. Through \(D\), the app creates line \(DE\parallel AC\), meeting ray \(\overrightarrow{BC}\) beyond \(C\) at \(E\). 3. Triangles \(ACD\) and \(ACE\) share base \(\overline{AC}\). 4. Their third vertices \(D\) and \(E\) lie on a line parallel to \(AC\), so the triangles have equal altitudes to \(AC\) and therefore equal areas. 5. The quadrilateral area is \([ABC]+[ACD]\). 6. Since \([ACD]=[ACE]\), this equals \([ABC]+[ACE]=[ABE]\).

The app creates \(DE\parallel AC\) with \(E\) on ray \(BC\). The equal-area triangles \(ACD\) and \(ACE\) show that \([ABCD]=[ABE]\).

- Translate each circle membership into one side-length equality. - Identify the quadrilateral whose opposite sides receive those equalities. - Check the vertex order before applying a quadrilateral classification theorem.

1. Since \(Q\) lies on the circle centered at \(P\) with radius \(AB\), \(PQ=AB\). 2. Since \(Q\) lies on the circle centered at \(B\) with radius \(AP\), \(BQ=AP\). 3. The side-of-line condition selects the intersection for which quadrilateral \(ABQP\) is simple rather than crossed. 4. In simple quadrilateral \(ABQP\), both pairs of opposite sides are congruent: \(AB=PQ\) and \(BQ=AP\). 5. Therefore, \(ABQP\) is a parallelogram. 6. Opposite sides of a parallelogram are parallel, so \(PQ\parallel AB\). Since \(\overline{AB}\) lies on \(\ell\), \(PQ\parallel\ell\).

The circle conditions give \(PQ=AB\) and \(BQ=AP\). Choosing \(Q\) opposite \(A\) across line \(BP\) makes \(ABQP\) a simple quadrilateral with both pairs of opposite sides congruent, so it is a parallelogram. Hence \(PQ\parallel\ell\). The other intersection can produce a crossed quadrilateral, so the orientation condition is necessary.

- Look for parallelograms created by pairs of parallel lines. - Use opposite sides of those parallelograms to compare the two parts of each outer side. - Relate the inner triangle to the medial triangle of the outer triangle.

1. The app creates the line through \(A\) parallel to \(BC\), the line through \(B\) parallel to \(AC\), and the line through \(C\) parallel to \(AB\). 2. Point \(A\) lies on \(\overline{YZ}\). Quadrilateral \(ABCY\) is a parallelogram, so \(AY=BC\). A second parallelogram gives \(AZ=BC\). Thus \(AY=AZ\), and \(A\) is the midpoint of \(\overline{YZ}\). 3. The same reasoning cyclically shows that \(B\) is the midpoint of \(\overline{ZX}\) and \(C\) is the midpoint of \(\overline{XY}\). 4. Therefore, \(\triangle ABC\) is the medial triangle of \(\triangle XYZ\). Each side of \(\triangle XYZ\) is twice the corresponding side of \(\triangle ABC\). 5. The scale factor from \(\triangle ABC\) to \(\triangle XYZ\) is \(2\), so the area scale factor is \(2^2=4\).

The app-created parallels make \(A\), \(B\), and \(C\) the midpoints of \(\overline{YZ}\), \(\overline{ZX}\), and \(\overline{XY}\), respectively. Thus \(\triangle ABC\) is the medial triangle of \(\triangle XYZ\), and \([XYZ]=4[ABC]\).

- A parallel cross-section creates a triangle similar to the original. - Determine the side-length scale factor whose square is \(\frac{1}{2}\). - A square diagonal can produce the needed irrational ratio exactly.

1. In the app, create a square whose side length is \(AB\). Its diagonal has length \(AB\sqrt{2}\). 2. Bisect that diagonal. The half-diagonal has length \(\frac{AB\sqrt{2}}{2}=\frac{AB}{\sqrt{2}}\). 3. Transfer this half-diagonal length from \(A\) along \(\overline{AB}\) to locate \(D\). Thus \(\frac{AD}{AB}=\frac{1}{\sqrt{2}}\). 4. Through \(D\), create the line parallel to \(BC\), and let it meet \(\overline{AC}\) at \(E\). 5. Because \(DE\parallel BC\), triangles \(ADE\) and \(ABC\) are similar with scale factor \(\frac{1}{\sqrt{2}}\). 6. Their area ratio is the square of the scale factor: \(\frac{[ADE]}{[ABC]}=\left(\frac{1}{\sqrt{2}}\right)^2=\frac{1}{2}\).

In the app, create a length \(\frac{AB}{\sqrt{2}}\) as half the diagonal of a square with side \(AB\). Mark \(D\) on \(\overline{AB}\) so that \(AD=\frac{AB}{\sqrt{2}}\), then create \(DE\parallel BC\). Similarity gives \([ADE]=\frac{1}{2}[ABC]\).

- If \(P\) is a midpoint, the endpoints are images under a half-turn about \(P\). - Determine where the image of line \(\ell\) lies after that half-turn. - Use the fact that nonparallel lines intersect exactly once.

1. The app creates the image \(\ell'\) of line \(\ell\) under a \(180^\circ\) rotation about \(P\). The image line satisfies \(\ell'\parallel\ell\). 2. Since \(\ell\) is not parallel to \(m\), line \(\ell'\) meets \(m\) at exactly one point \(B\). 3. The app reflects \(B\) across \(P\) to point \(A\), so \(P\) is the midpoint of \(\overline{AB}\). 4. Because \(B\) lies on the rotated image \(\ell'\), its half-turn image \(A\) lies on \(\ell\). 5. The unique intersection \(B=\ell'\cap m\) makes the segment unique.

The app rotates \(\ell\) by \(180^\circ\) about \(P\) to obtain \(\ell'\parallel\ell\). Let \(B=\ell'\cap m\), then reflect \(B\) across \(P\) to obtain \(A\). The result is unique because \(\ell'\) and \(m\) have exactly one intersection.