Free math worksheets

- Assume the circles share a point. - Express the distance from that point to the common center in two ways. - Compare the resulting radius values.

1. Let the circles have common center \(O\) and different radii \(r_1\) and \(r_2\). Assume for contradiction that they share a point \(P\). 2. Since \(P\) lies on the first circle, \(OP=r_1\). 3. Since \(P\) also lies on the second circle, \(OP=r_2\). 4. Therefore, \(r_1=r_2\), contradicting the fact that the radii are different. 5. Hence the two concentric circles have no common point.

A common point would have to be both radii away from the shared center, forcing the radii to be equal. This contradiction shows that the circles do not intersect.

- Begin by assuming the conclusion is false. - Focus on the angles each line makes with the same ray at \(A\). - Identify the uniqueness condition that the assumption would violate.

1. Assume for contradiction that \(m\) and \(n\) are distinct lines. 2. Because both lines pass through \(A\), they form an angle at \(A\). 3. Since \(m\perp\ell\) and \(n\perp\ell\), each line forms a \(90^\circ\) angle with the same ray of \(\ell\) at \(A\). 4. From a fixed ray at a point, there is only one ray on a given side that forms a \(90^\circ\) angle with it. Thus the rays of \(m\) and \(n\) on that side must coincide. 5. This contradicts the assumption that \(m\) and \(n\) are distinct. Therefore, \(m=n\).

Assuming \(m\ne n\) produces two different perpendiculars to \(\ell\) through the same point \(A\), contradicting the uniqueness of a \(90^\circ\) ray from a fixed ray at \(A\). Therefore, \(m\) and \(n\) are the same line.

- Let the fourth angle have measure \(x^\circ\), and translate “not a right angle” into a statement about \(x\). - Add that assumed angle measure to the total of the three known right angles. - Compare the resulting total with the quadrilateral angle-sum theorem.

1. Assume for contradiction that the fourth angle is not a right angle, and let its measure be \(x^\circ\). Then \(x\ne90\). 2. The three right angles have total measure \(3\cdot90^\circ=270^\circ\). 3. Under the assumption, the quadrilateral's angle sum would be \((270+x)^\circ\), which cannot equal \(360^\circ\) because \(x\ne90\). 4. This contradicts the theorem that the interior angles of every quadrilateral sum to \(360^\circ\). 5. Therefore, the assumption is false, so the fourth angle must be a right angle.

If the fourth angle had measure \(x^\circ\) with \(x\ne90\), the total angle measure would be \((270+x)^\circ\ne360^\circ\), contradicting the quadrilateral angle-sum theorem. Therefore, the fourth angle measures \(90^\circ\).

- Separate the existence of a midpoint from the claim that it is unique. - Assume there are two different points with the midpoint property. - Compare both points' distances from the same endpoint along one ray.

1. A segment has a midpoint by the midpoint existence theorem, so it remains to prove that the midpoint is unique. 2. Assume for contradiction that \(\overline{AB}\) has two distinct midpoints, \(M\) and \(N\). 3. Since both are midpoints, \(AM=MB\) and \(AN=NB\). 4. Therefore, \(AM=\frac{1}{2}AB\) and \(AN=\frac{1}{2}AB\), so \(AM=AN\). 5. Both \(M\) and \(N\) lie on ray \(\overrightarrow{AB}\). On a ray, there is only one point at a given positive distance from its endpoint. 6. Thus \(M=N\), contradicting the assumption that they are distinct. 7. Therefore, the existing midpoint of \(\overline{AB}\) is unique, so the segment has exactly one midpoint.

A midpoint of \(\overline{AB}\) exists. Assuming two distinct midpoints forces both points to lie on \(\overrightarrow{AB}\) at the same distance \(\frac{1}{2}AB\) from \(A\). A ray has only one point at a given positive distance from its endpoint, so the two points must coincide. Hence \(\overline{AB}\) has exactly one midpoint.

- Begin by assuming the prohibited configuration exists. - Use the defining measure condition for each angle type. - Compare the partial angle sum with the total available in a triangle.

1. Assume for contradiction that a triangle has two obtuse angles. 2. Each obtuse angle measures more than \(90^\circ\), so the sum of those two angles is greater than \(180^\circ\). 3. The three interior angles of a triangle have total measure exactly \(180^\circ\). 4. The two obtuse angles alone would exceed the entire allowed angle sum, which is impossible. 5. Therefore, a triangle cannot contain two obtuse angles.

Assuming two obtuse angles makes their sum greater than \(180^\circ\), contradicting the triangle angle-sum theorem. Therefore, a triangle can have at most one obtuse angle.

- Assume the tangent contains both the center and its point of tangency. - Determine where the radius to the tangency point would lie. - Compare that placement with the required radius-tangent relationship.

1. Assume for contradiction that tangent line \(m\) touches the circle at \(T\) and also passes through center \(O\). 2. Because \(O\) and \(T\) both lie on \(m\), radius \(\overline{OT}\) lies along line \(m\). 3. A tangent line is perpendicular to the radius at the point of tangency, so \(m\perp OT\). 4. This would require line \(m\) to be perpendicular to itself, which is impossible. 5. Therefore, a tangent line cannot pass through the circle's center.

Assuming a tangent contains the center makes the radius to the tangency point lie on the tangent, while the tangent-radius theorem requires the same two lines to be perpendicular. This contradiction proves that no tangent passes through the center.

- Assume the two lines have the relationship opposite the conclusion. - Identify what point would exist under that assumption and why it is not on \(\ell\). - Apply the uniqueness condition for a parallel through one point.

1. Assume for contradiction that \(m\) and \(n\) are not parallel. 2. Since they are distinct coplanar lines and are not parallel, they intersect at some point \(P\). 3. Because \(m\parallel\ell\), line \(m\) does not meet \(\ell\). Since \(P\) lies on \(m\), \(P\) does not lie on \(\ell\). 4. Both \(m\) and \(n\) pass through \(P\) and are parallel to \(\ell\). 5. Through a point not on a line, there is exactly one line parallel to the given line. Thus \(m\) and \(n\) would have to be the same line, contradicting the fact that they are distinct. 6. Therefore, \(m\parallel n\).

If \(m\) and \(n\) intersected, their intersection point would not lie on \(\ell\) and would have two distinct lines through it parallel to \(\ell\), contradicting the uniqueness of a parallel through a point. Hence \(m\parallel n\).

- Begin by assuming the scalene triangle has the prohibited angle relationship. - Connect equal angle measures to the sides opposite them. - Compare the resulting side classification with the original triangle type.

1. Assume for contradiction that a scalene triangle has two congruent angles. 2. In a triangle, congruent angles have congruent opposite sides. 3. Therefore, the two sides opposite the congruent angles are congruent. 4. The triangle would then be isosceles, contradicting the assumption that it is scalene. 5. Therefore, a scalene triangle cannot have two congruent angles.

If a scalene triangle had two congruent angles, their opposite sides would be congruent, making the triangle isosceles. This contradiction proves that all three angles of a scalene triangle must have different measures.

- Combine the rhombus condition with the assumption that the figure is not a square to infer something about its angles. - Use the diagonal test for a rectangle that applies to parallelograms. - Compare the resulting angle information with the consequence of the assumption.

1. Assume for contradiction that a rhombus has congruent diagonals but is not a square. 2. Because the quadrilateral is a rhombus, all four sides are congruent. Under the assumption that it is not a square, at least one interior angle is not a right angle; otherwise, four congruent sides and four right angles would make it a square. 3. Every rhombus is a parallelogram. 4. A parallelogram with congruent diagonals is a rectangle, so all four of its interior angles are right angles. 5. This contradicts the consequence of the assumption that at least one angle is not a right angle. 6. Therefore, the rhombus must be a square.

Assume the rhombus is not a square. Since its four sides are already congruent, at least one angle would have to be nonright. But a rhombus is a parallelogram, and congruent diagonals make that parallelogram a rectangle, so all four angles are right. This contradiction proves the rhombus is a square.

- Translate “not a rhombus” into a statement about at least one side length. - Use both pairs of congruent opposite sides in the parallelogram. - Compare every side with the given congruent consecutive pair.

1. Let parallelogram \(ABCD\) have \(AB=BC\). Assume for contradiction that \(ABCD\) is not a rhombus. 2. Since a rhombus has four congruent sides, the assumption means at least one of \(BC\), \(CD\), or \(AD\) is not congruent to \(AB\). 3. Opposite sides of a parallelogram are congruent, so \(AB=CD\) and \(BC=AD\). 4. Together with \(AB=BC\), these equalities give \(AB=BC=CD=AD\). 5. Thus none of the other three sides differs from \(AB\), contradicting the consequence of the assumption in step 2. 6. Therefore, \(ABCD\) is a rhombus.

Assume the parallelogram is not a rhombus, so at least one side differs in length from \(AB\). But \(AB=BC\), and opposite sides of a parallelogram give \(AB=CD\) and \(BC=AD\). Hence all four sides equal \(AB\), contradicting the assumption. Therefore, the parallelogram is a rhombus.

- Negate the inequality you are trying to prove. - Express the exterior angle using the two remote interior angles. - Check what the assumed inequality would force for the remaining remote angle.

1. Assume for contradiction that \(m\angle ACD\le m\angle A\). 2. By the exterior angle theorem, \(m\angle ACD=m\angle A+m\angle B\). 3. Substituting gives \(m\angle A+m\angle B\le m\angle A\), so \(m\angle B\le0^\circ\). 4. An angle of a nondegenerate triangle must have positive measure, so this is impossible. 5. Therefore, \(m\angle ACD>m\angle A\).

Assuming \(m\angle ACD\le m\angle A\) forces \(m\angle B\le0^\circ\), which is impossible. Hence \(m\angle ACD>m\angle A\).

- Assume three angles are congruent and the fourth is different. - Use the supplementary relationship between opposite angles in a cyclic quadrilateral. - Determine what the first opposite pair forces, then examine the second pair.

1. Assume for contradiction that cyclic quadrilateral \(ABCD\) has \(\angle A\cong\angle B\cong\angle C\), but \(\angle D\) is not congruent to them. 2. Opposite angles of a cyclic quadrilateral are supplementary, so \(m\angle A+m\angle C=180^\circ\). 3. Since \(\angle A\cong\angle C\), each has measure \(90^\circ\). Therefore, \(m\angle B=90^\circ\) as well. 4. Because \(\angle B\) and \(\angle D\) are supplementary, \(m\angle D=90^\circ\). 5. This makes all four angles congruent, contradicting the assumption that exactly three are congruent. 6. Therefore, a cyclic quadrilateral cannot have exactly three congruent interior angles.

Assuming exactly three congruent angles forces each of them to be \(90^\circ\), and then the fourth angle must also be \(90^\circ\). This contradiction proves the claim.

- First identify a line through the point that is tangent to the circle. - For uniqueness, assume two distinct tangent lines exist at the same point. - Relate each tangent to the radius and use uniqueness of a perpendicular through a point.

1. Let \(T\) be a point on a circle with center \(O\). The line through \(T\) perpendicular to \(\overline{OT}\) is tangent to the circle at \(T\), so a tangent at \(T\) exists. 2. To prove uniqueness, assume for contradiction that two distinct lines \(t_1\) and \(t_2\) are tangent at \(T\). 3. A tangent at \(T\) is perpendicular to radius \(\overline{OT}\), so both \(t_1\perp OT\) and \(t_2\perp OT\). 4. Through a given point, there is exactly one line perpendicular to a given line. 5. Therefore, \(t_1=t_2\), contradicting the assumption that they are distinct. 6. Hence there is exactly one tangent line at \(T\).

The line through \(T\) perpendicular to \(\overline{OT}\) is tangent to the circle, so a tangent exists. Two alleged tangents at \(T\) would both be perpendicular to \(\overline{OT}\). Since only one perpendicular to \(\overline{OT}\) passes through \(T\), the two lines would coincide. Therefore, the tangent is unique.

- Turn \(AB\ne AC\) into one of two strict length inequalities. - Write a Pythagorean equation for each right triangle formed by the altitude. - Use the midpoint condition to compare the two hypotenuse squares with the assumed inequality.

1. Assume for contradiction that \(AB\ne AC\). Then either \(AB>AC\) or \(AC>AB\). 2. Since \(\overline{AD}\) is an altitude, triangles \(ADB\) and \(ADC\) are right triangles. 3. Since \(D\) is the midpoint of \(\overline{BC}\), \(BD=DC\), and the triangles share leg \(AD\). 4. The Pythagorean theorem gives \(AB^2=AD^2+BD^2\) and \(AC^2=AD^2+DC^2\). 5. Because \(BD=DC\), these equations give \(AB^2=AC^2\), so \(AB=AC\). 6. This contradicts both possible inequalities from the assumption. Therefore, \(AB=AC\).

Assume \(AB\ne AC\), so one of the two lengths is greater. The two right triangles share leg \(AD\) and have \(BD=DC\). Thus \(AB^2=AD^2+BD^2=AD^2+DC^2=AC^2\), which forces \(AB=AC\), contradicting the assumption. Therefore, \(AB=AC\).

- Assume the diagonals do bisect each other. - Identify the quadrilateral test triggered by that diagonal property. - Compare the resulting number of parallel side pairs with the definition given.

1. Assume for contradiction that a trapezoid with exactly one pair of parallel opposite sides has diagonals that bisect each other. 2. A quadrilateral whose diagonals bisect each other is a parallelogram. 3. A parallelogram has two pairs of parallel opposite sides. 4. This contradicts the definition that the trapezoid has exactly one pair of parallel opposite sides. 5. Therefore, the diagonals of such a trapezoid cannot bisect each other.

If the diagonals bisected each other, the quadrilateral would be a parallelogram and would have two pairs of parallel sides. That contradicts the stated trapezoid definition, so its diagonals cannot bisect each other.

- Assume the nonsquare rhombus is cyclic. - Compare the opposite-angle properties of parallelograms and cyclic quadrilaterals. - Determine what congruent supplementary angles must measure.

1. Assume for contradiction that a nonsquare rhombus is cyclic. 2. Opposite angles of a rhombus are congruent because a rhombus is a parallelogram. 3. Opposite angles of a cyclic quadrilateral are supplementary. 4. Two angles that are both congruent and supplementary must each measure \(90^\circ\). 5. Thus the rhombus has right angles and is a square, contradicting the assumption that it is not a square. 6. Therefore, a nonsquare rhombus cannot be inscribed in a circle.

A cyclic rhombus would have opposite angles that are both congruent and supplementary, forcing right angles. It would therefore be a square, contradicting the nonsquare condition.

- Assume the line has two different tangency points. - Consider the radii to both alleged tangency points. - Use the uniqueness of the perpendicular from a point to a line.

1. Assume for contradiction that line \(t\) is tangent to a circle with center \(O\) at two distinct points \(A\) and \(B\). 2. A radius to a point of tangency is perpendicular to the tangent, so \(OA\perp t\) and \(OB\perp t\). 3. From point \(O\), there is exactly one perpendicular segment to line \(t\). 4. Therefore, the perpendicular feet \(A\) and \(B\) must be the same point. 5. This contradicts the assumption that \(A\) and \(B\) are distinct. Hence a line cannot be tangent to one circle at two distinct points.

Two tangency points would both be the foot of the perpendicular from the center to the same line. That perpendicular foot is unique, so the points would coincide. Therefore, two distinct tangency points are impossible.

- First intersect two internal angle bisectors and use their distance properties to establish existence. - For uniqueness, assume there are two different incenters. - Identify the same two lines on which every incenter must lie.

1. The internal angle bisectors of \(\angle A\) and \(\angle B\) intersect at a point \(I\) inside triangle \(ABC\). 2. Because \(I\) lies on the bisector of \(\angle A\), its perpendicular distances to lines \(AB\) and \(AC\) are equal. Because \(I\) lies on the bisector of \(\angle B\), its distances to lines \(BA\) and \(BC\) are equal. 3. Thus \(I\) is equidistant from all three sides, so an incenter exists. 4. To prove uniqueness, assume for contradiction that the triangle has two distinct incenters, \(I\) and \(J\). 5. Every incenter lies on the internal angle bisectors of \(\angle A\) and \(\angle B\), so both \(I\) and \(J\) lie at the intersection of those same two lines. 6. Two nonparallel lines intersect at exactly one point, so \(I=J\), contradicting the assumption that the incenters are distinct. 7. Hence a triangle has exactly one incenter.

The internal angle bisectors of \(\angle A\) and \(\angle B\) intersect at a point equidistant from all three sides, so an incenter exists. Any incenter must lie at that same unique intersection, so two distinct incenters are impossible.

- Write the side line perpendicular to each altitude. - Recall what happens to two lines perpendicular to parallel lines. - Compare that conclusion with the fact that two triangle sides meet at a vertex.

1. Assume for contradiction that the altitudes from vertices \(A\) and \(B\) are parallel. 2. The altitude from \(A\) is perpendicular to line \(BC\), and the altitude from \(B\) is perpendicular to line \(AC\). 3. If the two altitudes are parallel, then lines \(BC\) and \(AC\), each perpendicular to one of those parallel lines, must be parallel. 4. But lines \(BC\) and \(AC\) meet at vertex \(C\), so they cannot be distinct parallel lines. 5. This contradiction shows that the two altitudes cannot be parallel.

Parallel altitudes would force the two opposite side lines to be parallel, even though they meet at a triangle vertex. Therefore, two triangle altitudes cannot be parallel.

- Assume that two distinct perpendicular bisectors exist. - Identify the point through which both lines must pass. - Use the uniqueness of a perpendicular through a point on a line.

1. A segment has one midpoint \(M\), and the line through \(M\) perpendicular to the segment is a perpendicular bisector. 2. Assume for contradiction that the segment has two distinct perpendicular bisectors, \(p\) and \(q\). 3. Both \(p\) and \(q\) must pass through the same midpoint \(M\), and both must be perpendicular to the line containing the segment. 4. Through a given point on a line, only one line can be perpendicular to that line. Therefore, \(p=q\), contradicting the assumption that they are distinct. 5. Hence a nonzero segment has exactly one perpendicular bisector.

Two supposed perpendicular bisectors would both pass through the segment's unique midpoint and be perpendicular to the same line. The perpendicular through that midpoint is unique, so the two lines would be identical. Therefore, the segment has exactly one perpendicular bisector.

- Consider the triangle formed by the chord endpoints and the center. - Assume the chord exceeds the diameter. - Apply the triangle inequality to the two radii and the chord.

1. Let \(\overline{AB}\) be any chord of a circle with center \(O\) and radius \(r\). 2. Assume for contradiction that \(AB>2r\), the length of a diameter. 3. In triangle \(AOB\), \(OA=OB=r\). The triangle inequality gives \(AB\le OA+OB=2r\). 4. This contradicts the assumption \(AB>2r\). 5. Therefore, no chord can be longer than a diameter. Equality occurs only when \(A\), \(O\), and \(B\) are collinear, so the chord itself is a diameter.

A chord longer than \(2r\) would violate the triangle inequality in \(\triangle AOB\), whose other two sides are radii. Therefore, every chord has length at most the diameter.

- Assume the incenter lies on one side. - Translate that location into a perpendicular distance. - Use the defining equal-distance property of the incenter.

1. Assume for contradiction that the incenter \(I\) of triangle \(ABC\) lies on side \(\overline{AB}\). 2. The perpendicular distance from \(I\) to line \(AB\) is then \(0\). 3. An incenter is equidistant from all three side lines, so the distances from \(I\) to lines \(AC\) and \(BC\) must also be \(0\). 4. Therefore, \(I\) would lie on \(AB\), \(AC\), and \(BC\) at the same time. 5. In a nondegenerate triangle, the three side lines do not pass through one common point. This is a contradiction. 6. Hence the incenter cannot lie on a side of a nondegenerate triangle.

If the incenter lay on a side, its distance to that side would be \(0\). Equal distances to all three sides would then force it onto all three side lines, which is impossible for a nondegenerate triangle.

- Translate “inside the circle” into a comparison involving the radius. - Use the radius at the alleged tangency point to identify the distance from the center to the line. - Compare that shortest distance with \(OP\), since \(P\) lies on the line.

1. Let the circle have center \(O\) and radius \(r\), and let \(P\) be inside it, so \(OP<r\). 2. Assume for contradiction that a line \(\ell\) through \(P\) is tangent to the circle at \(T\). 3. A radius to a point of tangency is perpendicular to the tangent line, so \(OT\perp\ell\). Therefore, \(OT\) is the perpendicular distance from \(O\) to \(\ell\), and \(OT=r\). 4. Because \(P\) lies on \(\ell\), the shortest distance from \(O\) to \(\ell\) cannot exceed \(OP\). Thus \(r=OT\le OP\). 5. This contradicts \(OP<r\). Therefore, no tangent line through an interior point exists.

A supposed tangent through \(P\) would be at perpendicular distance \(r\) from the center. Since \(P\) lies on that line, its distance from the center would satisfy \(OP\ge r\), contradicting \(OP<r\).

- Assume that one point and its translated image have identical coordinates. - Compare each coordinate before and after the translation. - Use the translation vector to find the displacement length.

1. Assume for contradiction that some point \(P(x, y)\) is fixed by the translation. 2. The image of \(P\) is \(P'(x+5, y-2)\). 3. If \(P\) were fixed, then \(P'=P\), so \(x+5=x\) and \(y-2=y\). 4. These equations would require \(5=0\) and \(-2=0\), which is impossible. 5. Therefore, the translation has no fixed point. 6. The displacement length is \(\sqrt{5^2+(-2)^2}=\sqrt{29}\), so every point moves \(\sqrt{29}\) units.

The translation has no fixed point, and every point moves \(\sqrt{29}\) units.

- Assume the two perpendicular bisectors have the relationship the problem says is impossible. - Translate each perpendicular-bisector condition into a relation with a side of the triangle. - Look for a contradiction involving the two different sides through vertex \(A\).

1. Assume for contradiction that \(p\parallel q\). 2. Since \(p\) is the perpendicular bisector of \(\overline{AB}\), \(AB\perp p\). 3. Since \(q\) is the perpendicular bisector of \(\overline{AC}\), \(AC\perp q\). Because \(p\parallel q\), this also gives \(AC\perp p\). 4. Thus both \(\overleftrightarrow{AB}\) and \(\overleftrightarrow{AC}\) pass through \(A\) and are perpendicular to \(p\). 5. There is only one line through \(A\) perpendicular to \(p\), so \(\overleftrightarrow{AB}=\overleftrightarrow{AC}\). Then \(A\), \(B\), and \(C\) are collinear. 6. This contradicts the fact that \(ABC\) is a nondegenerate triangle. Therefore, \(p\) and \(q\) cannot be parallel.

Assuming \(p\parallel q\) forces both \(\overleftrightarrow{AB}\) and \(\overleftrightarrow{AC}\) to be the unique line through \(A\) perpendicular to \(p\). That would make \(A\), \(B\), and \(C\) collinear, contradicting the existence of a nondegenerate triangle. Hence \(p\not\parallel q\).

- Determine where the center of any circle through two fixed points must lie. - Apply that condition to two different pairs of the three points. - Use the intersection behavior of the resulting lines to test whether two centers are possible.

1. The perpendicular bisectors of \(\overline{AB}\) and \(\overline{AC}\) are not parallel because \(A\), \(B\), and \(C\) are noncollinear. They intersect at one point, which is equidistant from \(A\), \(B\), and \(C\), so a circle through the three points exists. 2. Assume for contradiction that two distinct circles pass through \(A\), \(B\), and \(C\), with centers \(O_1\) and \(O_2\). 3. Each center is equidistant from \(A\) and \(B\), so both \(O_1\) and \(O_2\) lie on the perpendicular bisector of \(\overline{AB}\). 4. Each center is also equidistant from \(A\) and \(C\), so both lie on the perpendicular bisector of \(\overline{AC}\). 5. Two nonparallel lines have only one intersection point, so \(O_1=O_2\). 6. The two circles then have the same center and the same radius to \(A\), so they are the same circle, contradicting the assumption that they are distinct. 7. Therefore, exactly one circle passes through the three noncollinear points.

The perpendicular bisectors of \(\overline{AB}\) and \(\overline{AC}\) meet at a point equidistant from \(A\), \(B\), and \(C\), so a circle through the three points exists. Any such circle must have its center at that unique intersection. Two supposed circles would therefore have the same center and radius, contradicting their being distinct. Hence exactly one circle passes through the points.

- Assume the two bisectors meet at a right angle. - Examine the smaller triangle formed by the two vertices and the bisectors’ intersection. - Compare its angle sum with the angle sum of the original triangle.

1. In triangle \(ABC\), let the internal bisectors of \(\angle A\) and \(\angle B\) meet at \(I\). Assume for contradiction that the bisectors are perpendicular, so \(\angle AIB=90^\circ\). 2. In triangle \(AIB\), the other two angles have measures \(\frac{A}{2}\) and \(\frac{B}{2}\). 3. The triangle angle sum gives \(\frac{A}{2}+\frac{B}{2}+90^\circ=180^\circ\), so \(A+B=180^\circ\). 4. Then the angle sum of triangle \(ABC\) forces \(C=0^\circ\), which is impossible for a nondegenerate triangle. 5. Therefore, two internal angle bisectors of a nondegenerate triangle cannot be perpendicular.

Assuming the two internal bisectors are perpendicular forces the third angle of the triangle to be \(0^\circ\). This contradiction proves that the bisectors cannot be perpendicular.

- Negate the desired angle inequality carefully. - Separate the equality case from the strict-inequality case. - Relate the order of two angles to the order of their opposite sides.

1. Assume for contradiction that \(\angle C\le\angle B\). 2. If \(\angle C=\angle B\), then the opposite sides are congruent, so \(AB=AC\), contradicting \(AB>AC\). 3. If \(\angle C<\angle B\), then the side opposite \(\angle C\) is shorter than the side opposite \(\angle B\), so \(AB<AC\), again contradicting \(AB>AC\). 4. Both alternatives are impossible. Therefore, \(\angle C>\angle B\).

Assuming \(\angle C\le\angle B\) forces either \(AB=AC\) or \(AB<AC\), both of which contradict \(AB>AC\). Hence \(\angle C>\angle B\).

- Translate “not equilateral” into a statement about at least one pair of side lengths. - Apply the angle bisector theorem together with the midpoint condition for each median. - Compare the two side equalities with the consequence of the assumption.

1. Assume for contradiction that triangle \(ABC\) is not equilateral. Then at least one pair of its side lengths is unequal. 2. Let the median from \(A\) meet \(\overline{BC}\) at \(D\). Since \(BD=DC\) and \(\overline{AD}\) bisects \(\angle A\), the angle bisector theorem gives \(AB=AC\). 3. Let the median from \(B\) meet \(\overline{AC}\) at \(E\). Since \(AE=EC\) and \(\overline{BE}\) bisects \(\angle B\), the angle bisector theorem gives \(BA=BC\). 4. Therefore, \(AB=AC=BC\), so no pair of side lengths is unequal. 5. This contradicts the consequence of the assumption in step 1. Hence triangle \(ABC\) is equilateral.

Assume the triangle is not equilateral, so at least one pair of sides is unequal. The median-angle-bisector from \(A\) gives \(AB=AC\), and the one from \(B\) gives \(AB=BC\). Thus all three sides are congruent, contradicting the assumption. Therefore, the triangle is equilateral.

- Express “not equilateral” as a consequence involving unequal angle measures. - Combine the angle-bisector property of the incenter with the equal-radius property of the circumcenter. - Repeat the comparison for a second pair of vertices and compare the result with the assumption.

1. Assume for contradiction that triangle \(ABC\) has a common incenter and circumcenter \(O\) but is not equilateral. Then at least two of its angles have different measures. 2. Since \(O\) is the incenter, \(AO\) and \(BO\) bisect \(\angle A\) and \(\angle B\). 3. Since \(O\) is the circumcenter, \(OA=OB\). Thus triangle \(AOB\) is isosceles, so \(\angle OAB=\angle ABO\). 4. These angles equal \(\frac{1}{2}\angle A\) and \(\frac{1}{2}\angle B\), so \(\angle A=\angle B\). 5. Applying the same argument to another pair of vertices gives \(\angle B=\angle C\). Hence \(\angle A=\angle B=\angle C\). 6. This contradicts the consequence of the assumption that at least two angles have different measures. Therefore, the triangle is equilateral.

Assume the triangle is not equilateral, so at least two angles differ. The common center lies on every angle bisector and is equidistant from the vertices. Each center triangle is therefore isosceles, which gives \(\angle A=\angle B=\angle C\). This contradicts the assumption, so the triangle is equilateral.

- Turn \(AB\ne AC\) into one of two strict inequalities. - Multiply that inequality by the common positive altitude factor used in the area formula. - Compare the two resulting expressions as areas of the same triangle.

1. Let the common altitude length be \(h>0\). Assume for contradiction that \(AB\ne AC\). Then either \(AB>AC\) or \(AC>AB\). 2. If \(AB>AC\), multiplying by the positive number \(\frac{h}{2}\) gives \(\frac{1}{2}ABh>\frac{1}{2}ACh\). If \(AC>AB\), the reverse strict inequality follows. 3. But \(\frac{1}{2}ABh\) and \(\frac{1}{2}ACh\) are both expressions for the area of the same triangle, using the two equal altitudes. 4. The same area cannot be both greater and less than itself. Thus either inequality from the assumption is impossible. 5. Therefore, \(AB=AC\).

Assume \(AB\ne AC\), so one side is longer. Multiplying the strict inequality by the common positive factor \(\frac{h}{2}\) would make one of the area expressions \(\frac{1}{2}ABh\) and \(\frac{1}{2}ACh\) larger than the other. Both equal the area of \(\triangle ABC\), a contradiction. Therefore, \(AB=AC\).

- Negate the equality that expresses “equidistant.” - If two positive distances are unequal, name the larger one and compare the associated right triangles. - Use the Pythagorean theorem to relate each half-chord to its distance from the center.

1. The conclusion to be proved is \(OM=ON\), so the contradiction assumption is \(OM\ne ON\). 2. Without loss of generality, suppose \(OM>ON\). 3. A perpendicular from the center to a chord bisects the chord, so \(AM=\frac{AB}{2}\) and \(CN=\frac{CD}{2}\). 4. Right triangles \(\triangle OMA\) and \(\triangle ONC\) have equal hypotenuse lengths because \(OA=OC=r\). 5. By the Pythagorean theorem, \(AM^2=r^2-OM^2\) and \(CN^2=r^2-ON^2\). 6. Since \(OM>ON\), it follows that \(OM^2>ON^2\), so \(AM^2<CN^2\) and therefore \(AM<CN\). 7. Thus \(AB=2AM<2CN=CD\), contradicting the given fact that \(AB=CD\). 8. Therefore, \(OM=ON\), and the congruent chords are equidistant from the center.

a) Assume for contradiction that \(OM\ne ON\). b) If, without loss of generality, \(OM>ON\), then \(AM^2=r^2-OM^2<r^2-ON^2=CN^2\). Hence \(AB=2AM<2CN=CD\), contradicting \(AB=CD\). Therefore, \(OM=ON\).

- Let the perpendicular distances from the center to the two chords be \(OM\) and \(ON\). - Express each half-chord using a radius and the center-to-chord distance. - Assume the longer chord is not closer and compare the resulting half-lengths.

1. Let \(M\) and \(N\) be the feet of the perpendiculars from \(O\) to \(\overline{AB}\) and \(\overline{CD}\), respectively. These perpendiculars bisect the chords. 2. Assume for contradiction that \(AB\) is not closer to \(O\), so \(OM\ge ON\). 3. In right triangles \(OMA\) and \(ONC\), the hypotenuses \(OA\) and \(OC\) are equal radii. 4. From the Pythagorean theorem, \(AM^2=OA^2-OM^2\) and \(CN^2=OC^2-ON^2\). Since \(OM\ge ON\), it follows that \(AM^2\le CN^2\), so \(AM\le CN\). 5. Therefore, \(AB=2AM\le2CN=CD\), contradicting \(AB>CD\). 6. Hence \(OM<ON\), so the longer chord \(\overline{AB}\) is closer to the center.

Assuming the longer chord is at least as far from the center forces its half-length to be no greater than the other chord's half-length, contradicting \(AB>CD\). Therefore, the longer chord is closer to the center.

- Assume the median is also perpendicular to the side it meets. - Identify the special line created by a midpoint and a perpendicular. - Combine the resulting equal lengths with the given isosceles condition.

1. Let \(D\) be the midpoint of \(\overline{AC}\), so \(BD\) is the median from \(B\). 2. Assume for contradiction that \(BD\) is also an altitude. Then \(BD\perp AC\). 3. Since \(D\) is the midpoint of \(AC\) and \(BD\perp AC\), line \(BD\) is the perpendicular bisector of \(\overline{AC}\). 4. Point \(B\) lies on this perpendicular bisector, so \(BA=BC\). 5. The given condition is \(BA=AC\). Therefore, \(BA=BC=AC\), making the triangle equilateral. 6. This contradicts the statement that the triangle is not equilateral. Hence the median from \(B\) to \(AC\) cannot also be an altitude.

If the median from \(B\) were also perpendicular to \(AC\), it would be the perpendicular bisector of \(AC\), giving \(BA=BC\). Together with \(BA=AC\), this would make the triangle equilateral, a contradiction.

- Check whether each intermediate statement depends on the opening assumption. - A redundant contradiction assumption can be removed without changing a direct proof. - To make the assumption useful, apply the contrapositive of the diameter-angle theorem before using the equal distances from \(M\).

1. The argument is logically valid, but the assumption \(\angle A\ne90^\circ\) is not used to derive any intermediate fact. Deleting the assumption and the final contradiction leaves a complete direct proof. 2. For a genuine indirect proof, assume \(\angle A\ne90^\circ\). 3. By the contrapositive of the theorem that an angle subtending a diameter is a right angle, point \(A\) cannot lie on the circle with diameter \(\overline{BC}\). 4. Since \(M\) is the midpoint of \(BC\), \(BM=CM=\frac{1}{2}BC\). The given condition also gives \(AM=\frac{1}{2}BC\). 5. Therefore, \(AM=BM=CM\), so \(A\), \(B\), and \(C\) lie on the circle centered at \(M\). 6. Because \(M\) lies on \(BC\), segment \(\overline{BC}\) is a diameter of that circle. Thus \(A\) does lie on the circle with diameter \(\overline{BC}\), contradicting Step 3. 7. Hence \(\angle A=90^\circ\).

a) The argument is logically valid, but it is a direct proof with an unused contradiction assumption: the circle-and-diameter reasoning proves \(\angle A=90^\circ\) without the first sentence. b) Assume \(\angle A\ne90^\circ\). Then, by the contrapositive of the diameter-angle theorem, \(A\) is not on the circle with diameter \(BC\). But \(AM=BM=CM\) places \(A\), \(B\), and \(C\) on the circle centered at \(M\), whose diameter is \(BC\). This contradiction proves \(\angle A=90^\circ\).

- Turn \(AB\ne AC\) into one strict inequality by naming the longer side. - Use the shared altitude in two Pythagorean equations to compare \(BD\) and \(CD\). - Use the equal bisected angles and the two right angles to obtain a conflicting comparison.

1. Assume for contradiction that \(AB\ne AC\). Without loss of generality, suppose \(AB>AC\). 2. Since \(AD\perp BC\), triangles \(ABD\) and \(ACD\) are right triangles sharing leg \(AD\). 3. By the Pythagorean theorem, \(BD^2=AB^2-AD^2\) and \(CD^2=AC^2-AD^2\). 4. The assumption \(AB>AC\) gives \(AB^2>AC^2\), so \(BD^2>CD^2\) and therefore \(BD>CD\). 5. However, \(\angle BAD=\angle DAC\) because \(AD\) bisects \(\angle A\), and \(\angle ADB=\angle ADC=90^\circ\). Thus \(\triangle ABD\sim\triangle ACD\) by AA. 6. Side \(AD\) corresponds to itself, so the similarity scale factor is \(1\). Therefore, \(BD=CD\). 7. This contradicts \(BD>CD\). Hence \(AB=AC\).

Assume, without loss of generality, that \(AB>AC\). The Pythagorean theorem then gives \(BD^2=AB^2-AD^2>AC^2-AD^2=CD^2\), so \(BD>CD\). But the two right triangles are AA-similar, and their shared corresponding side \(AD\) makes the scale factor \(1\), forcing \(BD=CD\). This contradiction proves \(AB=AC\).

- First rule out coincident centers using one shared point. - Apply the center-tangency-point alignment at both alleged tangency points. - Determine what two shared points on the line of centers mean for both circles.

1. Assume for contradiction that distinct circles with centers \(O_1\) and \(O_2\) are tangent at two different points \(A\) and \(B\). 2. The centers must be distinct. If \(O_1=O_2\), the shared point \(A\) would give both circles the same radius, so the circles would be identical. 3. At a point of tangency, the two centers and the tangency point are collinear. Therefore, \(O_1\), \(O_2\), and \(A\) are collinear, and \(O_1\), \(O_2\), and \(B\) are collinear. 4. Thus, both \(A\) and \(B\) lie on line \(O_1O_2\). 5. For either circle, two distinct points on the circle and on a line through its center are the endpoints of a diameter. Hence \(A\) and \(B\) are diameter endpoints for both circles. 6. Both circles therefore have the same midpoint of \(\overline{AB}\) as center and the same radius \(\frac{1}{2}AB\), so they are the same circle. 7. This contradicts the assumption that the circles are distinct. Therefore, two distinct circles cannot be tangent at two different points.

Assuming two tangency points forces both circles to have the same diameter \(\overline{AB}\), hence the same center and radius. That contradicts their being distinct.

- Form three smaller triangles using the interior point. - Write one triangle inequality for each side of the original triangle. - Add the inequalities and compare the result with the assumption.

1. Assume for contradiction that \(PA+PB+PC\le\frac{1}{2}(AB+BC+CA)\). 2. The triangle inequalities in triangles \(APB\), \(BPC\), and \(CPA\) give \(AB<PA+PB\), \(BC<PB+PC\), and \(CA<PC+PA\). 3. Adding these inequalities gives \(AB+BC+CA<2(PA+PB+PC)\). 4. Dividing by \(2\) gives \(\frac{1}{2}(AB+BC+CA)<PA+PB+PC\). 5. This contradicts the assumed inequality. Therefore, the stated strict inequality holds.

The three triangle inequalities add to \(AB+BC+CA<2(PA+PB+PC)\), contradicting the assumption that the distance sum is at most half the perimeter.

- Double the median by reflecting its endpoint across the side midpoint. - Identify the parallelogram created by the two bisecting diagonals. - Apply the triangle inequality to the new triangle.

1. Assume for contradiction that \(m_a\ge\frac{AB+AC}{2}\). 2. Reflect \(A\) across \(M\) to point \(A'\). Then \(AA'=2m_a\). 3. Since \(M\) is the midpoint of both \(\overline{BC}\) and \(\overline{AA'}\), quadrilateral \(ABA'C\) is a parallelogram. 4. Therefore, \(BA'=AC\). 5. In triangle \(ABA'\), the triangle inequality gives \(AA'<AB+BA'=AB+AC\). 6. Thus, \(2m_a<AB+AC\), or \(m_a<\frac{AB+AC}{2}\), contradicting the assumption.

Reflecting \(A\) across the midpoint of \(BC\) creates a triangle in which the triangle inequality gives \(2m_a<AB+AC\). Therefore, \(m_a<\frac{AB+AC}{2}\).

- Assume three distinct intersection points occur on one line. - Use equal radii to place the center on perpendicular-bisector loci. - Compare the two bisectors' directions and their intersection points with the original line.

1. Assume for contradiction that line \(\ell\) intersects a circle with center \(O\) at three distinct points \(A\), \(B\), and \(C\), in that order. 2. Since \(OA=OB\), point \(O\) lies on the perpendicular bisector \(p\) of \(\overline{AB}\). 3. Since \(OB=OC\), point \(O\) lies on the perpendicular bisector \(q\) of \(\overline{BC}\). 4. Segments \(\overline{AB}\) and \(\overline{BC}\) lie on the same line \(\ell\), so both \(p\) and \(q\) are perpendicular to \(\ell\). 5. Because \(p\) and \(q\) both pass through \(O\), the uniqueness of a perpendicular through a point implies that they are the same line. 6. That one line would meet \(\ell\) at both the midpoint of \(\overline{AB}\) and the midpoint of \(\overline{BC}\). These midpoints are distinct because \(A\), \(B\), and \(C\) are distinct and ordered on \(\ell\). 7. This is impossible because two distinct lines intersect at most once. Therefore, a line intersects a circle at no more than two points.

Three intersections would place the circle's center on the perpendicular bisectors of two collinear subsegments. Both bisectors pass through the center and are perpendicular to the original line, so uniqueness forces them to be the same line. That line would then intersect the original line at two distinct midpoints, which is impossible. Therefore, at most two intersections can occur.

- Assume all three median comparisons fail at once. - Convert each comparison using a median-length formula. - Add the resulting three side-length inequalities.

1. Let the side lengths be \(a\), \(b\), and \(c\), with corresponding medians \(m_a\), \(m_b\), and \(m_c\). 2. Assume for contradiction that \(m_a\le\frac{a}{2}\), \(m_b\le\frac{b}{2}\), and \(m_c\le\frac{c}{2}\). 3. The median formulas give \(4m_a^2=2b^2+2c^2-a^2\), with analogous equations for the other two medians. 4. From \(m_a\le\frac{a}{2}\), obtain \(b^2+c^2\le a^2\). Similarly, \(c^2+a^2\le b^2\) and \(a^2+b^2\le c^2\). 5. Adding the three inequalities gives \(2(a^2+b^2+c^2)\le a^2+b^2+c^2\). 6. This would force \(a^2+b^2+c^2\le0\), impossible for a nondegenerate triangle. Therefore, at least one median is longer than half its corresponding side.

Assuming all three medians are at most half their corresponding sides produces the impossible inequality \(2(a^2+b^2+c^2)\le a^2+b^2+c^2\). Hence at least one median is longer.

- Assume two side lengths are unequal and order them from shorter to longer. - Compare the squared lengths of the medians to those two sides. - Look for a positive difference that contradicts congruent medians.

1. Assume for contradiction that triangle \(ABC\) has three congruent medians but is not equilateral. 2. Let side \(c\) be longer than side \(a\). Let \(m_a\) and \(m_c\) be the medians to sides \(a\) and \(c\). 3. By the median-length formulas, \(m_a^2=\frac{2b^2+2c^2-a^2}{4}\) and \(m_c^2=\frac{2a^2+2b^2-c^2}{4}\). 4. Subtracting gives \(m_a^2-m_c^2=\frac{3(c^2-a^2)}{4}>0\). 5. Thus, \(m_a>m_c\), contradicting the assumption that all three medians are congruent. 6. Therefore, no two side lengths can differ, and the triangle is equilateral.

If one side were longer than another, the median to the shorter side would be longer than the median to the longer side. This contradicts three congruent medians, so all three sides are congruent.

- The common point is inside the triangle because it is the incenter; use this to identify the triangle as acute. - Recall the angle formed at an incenter by the bisectors from \(B\) and \(C\), and the angle formed at an orthocenter by the altitudes from those vertices. - Compare both formulas with \(120^\circ\) after assuming \(\angle A\) is greater than or less than \(60^\circ\).

1. Let the common orthocenter and incenter be \(P\). Because an incenter lies inside its triangle, \(P\) is inside the triangle, so the triangle is acute. 2. Assume for contradiction that the triangle is not equilateral. Then at least one angle is not \(60^\circ\); relabel so \(\angle A\ne60^\circ\). 3. Since \(P\) is the incenter, \(m\angle BPC=90^\circ+\frac{1}{2}m\angle A\). 4. Since \(P\) is also the orthocenter of an acute triangle, \(m\angle BPC=180^\circ-m\angle A\). 5. If \(m\angle A>60^\circ\), the incenter formula makes \(m\angle BPC>120^\circ\), while the orthocenter formula makes \(m\angle BPC<120^\circ\), which is impossible. 6. If \(m\angle A<60^\circ\), the incenter formula makes \(m\angle BPC<120^\circ\), while the orthocenter formula makes \(m\angle BPC>120^\circ\), which is also impossible. 7. Both alternatives contradict the assumption \(\angle A\ne60^\circ\). Therefore, every angle is \(60^\circ\), and the triangle is equilateral.

Assume some angle, say \(\angle A\), is not \(60^\circ\). The incenter identity gives \(m\angle BPC=90^\circ+\frac{1}{2}m\angle A\), while the orthocenter identity gives \(m\angle BPC=180^\circ-m\angle A\). If \(\angle A\) is greater than \(60^\circ\), these values lie on opposite sides of \(120^\circ\); the same happens in reverse if it is less than \(60^\circ\). This contradiction forces all angles to be \(60^\circ\).