Free Math Worksheets

- Draw the diagonal connecting the vertices where the pairs of congruent sides meet. - List the two given pairs of congruent sides. - Identify the side shared by the two triangles. - Use the congruence conclusion to compare corresponding angles.

1. Draw diagonal \(\overline{AC}\), dividing the kite into \(\triangle ABC\) and \(\triangle ADC\). 2. The triangles satisfy \(AB = AD\) and \(CB = CD\) by the definition of a kite. 3. They also share side \(\overline{AC}\), so \(AC = AC\). 4. Therefore, \(\triangle ABC \cong \triangle ADC\) by SSS. 5. Corresponding angles of congruent triangles are congruent, so \(\angle ABC = \angle ADC\).

Diagonal \(\overline{AC}\) creates two triangles with three pairs of congruent sides. By SSS, \(\triangle ABC \cong \triangle ADC\), so \(\angle ABC = \angle ADC\).

- Use the side and angle properties of a rectangle. - Mark the given congruent segments. - Identify two sides and their included angle in each triangle. - Corresponding parts of congruent triangles are congruent.

1. Opposite sides of a rectangle are congruent, so \(AD = CB\). 2. The angles \(\angle DAP\) and \(\angle BCQ\) are both right angles. 3. The problem gives \(AP = CQ\). 4. Therefore, \(\triangle APD \cong \triangle CQB\) by SAS. 5. Corresponding sides of congruent triangles are congruent, so \(PD = QB\).

a) \(\triangle APD \cong \triangle CQB\) by SAS because \(AD = CB\), \(AP = CQ\), and the included angles are right angles. b) \(PD = QB\).

- Which corresponding sides are congruent because of the triangle congruence? - Combine that result with the given side equality. - Which parallelogram test uses two pairs of congruent opposite sides?

1. Since \(\triangle ABM \cong \triangle CDK\), corresponding sides give \(AB = CD\). 2. The other pair of opposite sides satisfies \(BC = AD\) by the given information. 3. Both pairs of opposite sides of \(ABCD\) are congruent. Therefore, \(ABCD\) is a parallelogram.

Congruence gives \(AB = CD\), and the problem gives \(BC = AD\). Since both pairs of opposite sides are congruent, \(ABCD\) is a parallelogram.

- What does triangle congruence tell you about the four diagonal segments? - Which parallelogram test involves diagonals that bisect each other?

1. Corresponding sides of the congruent triangles give \(PM = RM\) and \(QM = SM\). 2. Therefore, \(M\) is the midpoint of both diagonals \(PR\) and \(QS\). The diagonals bisect each other. 3. A quadrilateral whose diagonals bisect each other is a parallelogram. Therefore, \(PQRS\) is a parallelogram.

The congruent triangles give \(PM = RM\) and \(QM = SM\), so the diagonals bisect each other. Therefore, \(PQRS\) is a parallelogram.

- Identify the vertical angles at \(Z\). - List the given equal angle and side. - Which criterion uses two angles and their included side?

1. Angles \(\angle KZL\) and \(\angle MZN\) are vertical angles, so they are congruent. 2. The givens are \(\angle LKZ = \angle NMZ\) and \(KZ = MZ\). 3. The equal side lies between the two pairs of equal angles. Therefore, \(\triangle KLZ \cong \triangle MNZ\) by ASA.

\(\triangle KLZ \cong \triangle MNZ\) by ASA.

- Identify the side shared by the triangles. - Match the two given angle pairs. - Determine whether the shared side is included between those angles.

1. The triangles share side \(\overline{AB}\), so \(AB = BA\). 2. The givens provide two pairs of congruent angles: \(\angle DAB = \angle CBA\) and \(\angle DBA = \angle CAB\). 3. In each triangle, the shared side lies between the two given angles. 4. Therefore, \(\triangle ABD \cong \triangle BAC\) by ASA.

\(\triangle ABD \cong \triangle BAC\) by ASA.

- Match the corresponding sides in the congruence statement. - Which quadrilateral theorem uses two pairs of congruent opposite sides?

1. From \(\triangle ABC \cong \triangle CDA\), corresponding sides are congruent: \(AB = CD\) and \(BC = DA\). 2. Both pairs of opposite sides of quadrilateral \(ABCD\) are congruent. 3. A quadrilateral with both pairs of opposite sides congruent is a parallelogram. Therefore, \(ABCD\) is a parallelogram.

Quadrilateral \(ABCD\) is a parallelogram because both pairs of opposite sides are congruent.

- Compare the three sides of the two triangles. - Include the shared diagonal. - Use corresponding angles after proving congruence. - State the definition of an angle bisector.

1. Consider \(\triangle ABC\) and \(\triangle ADC\). 2. The kite gives \(AB = AD\) and \(CB = CD\). 3. The triangles share side \(\overline{AC}\), so \(AC = AC\). 4. Therefore, \(\triangle ABC \cong \triangle ADC\) by SSS. 5. Corresponding angles are congruent, so \(\angle BAC = \angle DAC\). Thus, \(\overline{AC}\) bisects the angle at \(A\).

\(\triangle ABC \cong \triangle ADC\) by SSS, so \(\angle BAC = \angle DAC\). Therefore, \(\overline{AC}\) bisects the angle at \(A\).

- Draw the angle bisector from \(C\) to \(\overline{AB}\). - Compare the two angles created at \(C\). - Use the triangle angle sum to compare the angles at \(D\). - Identify the shared side between the two smaller triangles.

1. Draw the angle bisector of \(\angle C\), meeting \(\overline{AB}\) at \(D\). 2. By the definition of an angle bisector, \(\angle ACD = \angle BCD\). 3. Since \(\alpha = \beta\), the third angles in \(\triangle ACD\) and \(\triangle BCD\) are also congruent: \(\angle ADC = \angle BDC\). 4. Segment \(\overline{CD}\) is common to both triangles and is included between the two pairs of congruent angles. 5. Therefore, \(\triangle ACD \cong \triangle BCD\) by ASA. 6. Corresponding sides of congruent triangles are congruent, so \(AC = BC\). Thus, \(\triangle ABC\) is isosceles.

The angle bisector creates triangles \(ACD\) and \(BCD\). They have two pairs of congruent angles and common included side \(\overline{CD}\), so they are congruent by ASA. Therefore, \(AC = BC\).

- Translate each isosceles-triangle statement into a pair of congruent sides. - Compare triangles \(ABD\) and \(ACD\). - Identify their shared side. - Use the definition of an angle bisector after proving the triangles congruent.

1. Compare \(\triangle ABD\) and \(\triangle ACD\). 2. Since \(\triangle ABC\) is isosceles with base \(\overline{BC}\), \(AB = AC\). 3. Since \(\triangle DBC\) is isosceles with base \(\overline{BC}\), \(DB = DC\). 4. The triangles share side \(\overline{AD}\), so \(AD = AD\). 5. Therefore, \(\triangle ABD \cong \triangle ACD\) by SSS. 6. Corresponding angles are congruent, so \(\angle BAD = \angle DAC\). Thus, \(\overrightarrow{AD}\) bisects \(\angle BAC\).

\(\triangle ABD \cong \triangle ACD\) by SSS because \(AB = AC\), \(DB = DC\), and \(AD\) is common. Therefore, \(\angle BAD = \angle DAC\), so \(\overrightarrow{AD}\) is the angle bisector.

- How is each exterior angle related to its adjacent interior angle? - What condition on two interior angles proves a triangle is isosceles? - Which sides are opposite \(\angle A\) and \(\angle B\)?

1. Each exterior angle forms a linear pair with its adjacent interior angle. Therefore, \(m\angle A + m\angle 1 = 180^\circ\) and \(m\angle B + m\angle 4 = 180^\circ\). 2. Since \(m\angle 1 = m\angle 4\), subtracting equal measures from \(180^\circ\) gives \(m\angle A = m\angle B\). 3. By the converse of the isosceles triangle theorem, the sides opposite \(\angle A\) and \(\angle B\) are congruent. Therefore, \(BC = AC\), and \(\triangle ABC\) is isosceles.

Equal exterior angles have equal supplementary interior angles, so \(m\angle A = m\angle B\). By the converse of the isosceles triangle theorem, \(AC = BC\), so \(\triangle ABC\) is isosceles.

- Use the fact that diagonals of a parallelogram bisect each other. - Express \(SP\) and \(SQ\) as differences of segments. - Look for vertical angles at \(S\). - Prove both pairs of opposite sides of \(APCQ\) congruent.

1. The diagonals of a parallelogram bisect each other, so \(AS = SC\) and \(BS = SD\). 2. Since \(SP = BS - BP\), \(SQ = SD - DQ\), \(BS = SD\), and \(BP = DQ\), it follows that \(SP = SQ\). 3. Angles \(\angle ASP\) and \(\angle CSQ\) are vertical angles. Therefore, \(\triangle ASP \cong \triangle CSQ\) by SAS, and \(AP = CQ\). 4. Angles \(\angle ASQ\) and \(\angle CSP\) are vertical angles. Therefore, \(\triangle ASQ \cong \triangle CSP\) by SAS, and \(AQ = CP\). 5. Both pairs of opposite sides of quadrilateral \(APCQ\) are congruent. Therefore, \(APCQ\) is a parallelogram.

The original parallelogram gives \(AS = SC\) and \(BS = SD\). Together with \(BP = DQ\), this yields \(SP = SQ\). Two SAS congruence arguments give \(AP = CQ\) and \(AQ = CP\). Since both pairs of opposite sides are congruent, \(APCQ\) is a parallelogram.

- What angles are formed because \(BD\) is an altitude? - Which hypotenuses and legs are congruent in the two right triangles? - Use congruence to determine the two parts of the base. - Then apply the Pythagorean theorem.

1. Because \(BD\) is an altitude, \(\angle ADB\) and \(\angle CDB\) are right angles. 2. The right triangles have congruent hypotenuses, \(AB=BC=10\,\text{cm}\), and share leg \(BD\). Therefore, \(\triangle ABD\cong\triangle CBD\) by the Hypotenuse-Leg theorem. 3. Corresponding parts of congruent triangles are congruent, so \(AD=CD\). Since \(AC=12\,\text{cm}\), each segment is \(6\,\text{cm}\). 4. In \(\triangle ABD\), \(BD^2=10^2-6^2=64\), so \(BD=8\,\text{cm}\).

The triangles are congruent by the Hypotenuse-Leg theorem, and \(BD=8\,\text{cm}\).

- Identify the two right triangles formed by the altitudes. - Which sides are congruent because the original triangle is isosceles? - Look for a pair of angles at vertex \(B\).

1. Consider right triangles \(ABD\) and \(CBE\). 2. Because \(AD\) and \(CE\) are altitudes, \(\angle ADB = \angle CEB = 90^{\circ}\). 3. The angles \(\angle ABD\) and \(\angle CBE\) are the same angle at vertex \(B\). 4. Since \(AB = BC\), the hypotenuses are congruent. 5. The triangles are congruent by AAS. Therefore, corresponding sides \(AD\) and \(CE\) are congruent.

Triangles \(ABD\) and \(CBE\) are congruent by AAS, so \(AD = CE\).

- Use diagonal \(AC\) to create two triangles. - Which angle pair is congruent because \(AB \parallel CD\)? - Which triangle congruence theorem applies? - Use corresponding angles from the congruent triangles to prove the other pair of sides parallel.

1. Consider diagonal \(AC\). 2. Since \(AB \parallel CD\), \(\angle BAC \cong \angle ACD\) by the alternate interior angles theorem. 3. In \(\triangle ABC\) and \(\triangle CDA\), \(AB = CD\), \(AC\) is a common side, and the included angles are congruent. Therefore, \(\triangle ABC \cong \triangle CDA\) by SAS. 4. Corresponding angles give \(\angle BCA \cong \angle CAD\). By the converse of the alternate interior angles theorem, \(BC \parallel AD\). 5. Both pairs of opposite sides are parallel, so \(ABCD\) is a parallelogram.

Using diagonal \(AC\) creates two triangles that are congruent by SAS. Corresponding alternate interior angles then prove \(BC \parallel AD\). Together with \(AB \parallel CD\), this makes \(ABCD\) a parallelogram.

- What kind of triangles are formed by the altitude? - Which hypotenuses are congruent? - Which leg is shared by both triangles?

1. Triangles \(ACD\) and \(BCD\) are right triangles because \(CD\) is an altitude. 2. Their hypotenuses are congruent because \(AC = BC\). 3. They share leg \(CD\). 4. Therefore, \(\triangle ACD \cong \triangle BCD\) by the Hypotenuse–Leg theorem. 5. Corresponding sides are congruent, so \(AD = BD\).

\(\triangle ACD \cong \triangle BCD\) by HL; therefore, \(AD = BD\).

- Use alternate interior angles formed by the diagonal and each pair of parallel sides. - Identify the side between the two equal angles.

1. Since \(AB \parallel CD\), alternate interior angles satisfy \(\angle BAC = \angle DCA\). 2. Since \(BC \parallel AD\), alternate interior angles satisfy \(\angle BCA = \angle DAC\). 3. The triangles share the included side \(\overline{AC}\). 4. Therefore, \(\triangle ABC \cong \triangle CDA\) by ASA.

The two pairs of alternate interior angles are congruent, and \(AC\) is the included shared side. Therefore, \(\triangle ABC \cong \triangle CDA\) by ASA.

- Add \(DB\) to the equal lengths \(AD\) and \(BF\). - Rewrite the resulting sums as the full sides of the triangles. - Check whether the given angle is included between the equal sides.

1. Add \(DB\) to both sides of \(AD = BF\): \(AD + DB = BF + DB\). 2. Since \(AD + DB = AB\) and \(BF + DB = FD\), it follows that \(AB = FD\). 3. The other givens are \(AC = FE\) and \(\angle BAC = \angle DFE\). 4. Each given angle is included between the two corresponding equal sides. Therefore, \(\triangle ABC \cong \triangle FDE\) by SAS.

Segment addition gives \(AB = FD\). With \(AC = FE\) and \(\angle BAC = \angle DFE\), the triangles are congruent by SAS.

- Use the base angles of the isosceles triangle. - Identify the shared side. - Combine those facts with the given equal segments.

1. Since \(AC = BC\), the base angles are congruent: \(\angle CAB = \angle CBA\). 2. Because \(D\) lies on \(\overline{AC}\) and \(E\) lies on \(\overline{BC}\), \(\angle DAB = \angle EBA\). 3. The triangles share side \(\overline{AB}\), and \(AD = BE\) is given. 4. Therefore, \(\triangle ABE \cong \triangle BAD\) by SAS.

\(\triangle ABE \cong \triangle BAD\) by SAS.

- Identify the angle at \(C\) that belongs to both triangles. - Use the equal legs of the isosceles triangle. - Check whether the equal side is included between the two equal angles.

1. The given equal sides are \(AC = BC\). 2. Since \(E\) lies on \(\overline{BC}\) and \(D\) lies on \(\overline{AC}\), the angles at \(C\) are the same: \(\angle ACE = \angle BCD\). 3. The other given angle pair is \(\angle CAE = \angle CBD\). 4. The equal sides \(AC\) and \(BC\) are included between the corresponding equal angles. Therefore, \(\triangle ACE \cong \triangle BCD\) by ASA.

\(\triangle ACE \cong \triangle BCD\) by ASA.

- Relate each interior angle to its marked exterior angle. - Use the fact that supplements of congruent angles are congruent. - Determine whether the equal angle is included between the equal sides.

1. Angles \(\angle FKH\) and \(\angle PEH\) are supplementary to the congruent marked exterior angles. 2. Supplements of congruent angles are congruent, so \(\angle FKH = \angle PEH\). 3. The givens also provide \(FK = PE\) and \(KH = EH\). 4. Each equal angle is included between the corresponding equal sides. Therefore, \(\triangle FKH \cong \triangle PEH\) by SAS.

\(\triangle FKH \cong \triangle PEH\) by SAS.

- Compare the two triangles formed by \(\overline{BD}\). - Use the shared side and the given equal sides. - State what the angle bisector tells you.

1. Consider \(\triangle ABD\) and \(\triangle CBD\). 2. The triangles share side \(\overline{BD}\), and \(AD = CD\) is given. 3. Because \(\overline{BD}\) bisects \(\angle ADC\), \(\angle ADB = \angle CDB\). 4. Therefore, \(\triangle ABD \cong \triangle CBD\) by SAS. 5. Corresponding sides are congruent, so \(AB = CB\). Thus, \(\triangle ABC\) is isosceles.

\(\triangle ABD \cong \triangle CBD\) by SAS, so \(AB = CB\). Therefore, \(\triangle ABC\) is isosceles.

- Match the two given angle pairs. - Use the shared side with an angle-based congruence criterion. - Apply corresponding parts of congruent triangles.

1. In \(\triangle ABD\) and \(\triangle CBD\), two pairs of angles are congruent: \(\angle ADB = \angle CDB\) and \(\angle DAB = \angle DCB\). 2. The triangles share side \(\overline{BD}\), which is opposite the second pair of congruent angles. 3. Therefore, \(\triangle ABD \cong \triangle CBD\) by AAS. 4. Corresponding sides are congruent, so \(AB = CB\). Thus, \(\triangle ABC\) is isosceles.

\(\triangle ABC\) is isosceles because \(\triangle ABD \cong \triangle CBD\) by AAS, which gives \(AB = CB\).

- Use the angle-bisector information at \(B\). - Combine it with the given equal angles. - Identify the shared side of the two smaller triangles.

1. Since \(\overline{BE}\) bisects \(\angle ABC\), \(\angle ABE = \angle CBE\). 2. The other given angle pair is \(\angle BAE = \angle BCE\). 3. Triangles \(\triangle ABE\) and \(\triangle CBE\) share side \(\overline{BE}\), which is opposite the second pair of equal angles. 4. Therefore, \(\triangle ABE \cong \triangle CBE\) by AAS. 5. Corresponding sides are congruent, so \(AB = CB\). Thus, \(\triangle ABC\) is isosceles.

\(\triangle ABE \cong \triangle CBE\) by AAS, so \(AB = CB\). Therefore, \(\triangle ABC\) is isosceles.

- Use the midpoint and altitude information at \(D\). - Identify the shared side of the two smaller triangles.

1. Since \(D\) is the midpoint of \(\overline{AC}\), \(AD = CD\). 2. Because \(\overline{ED}\) is perpendicular to \(\overline{AC}\) and \(B\) lies on \(\overline{ED}\), \(\angle ADB = \angle CDB = 90^\circ\). 3. Triangles \(\triangle ABD\) and \(\triangle CBD\) share side \(\overline{BD}\). 4. Therefore, \(\triangle ABD \cong \triangle CBD\) by SAS. 5. Corresponding sides are congruent, so \(AB = CB\). Thus, \(\triangle ABC\) is isosceles.

\(\triangle ABD \cong \triangle CBD\) by SAS, so \(AB = CB\). Therefore, \(\triangle ABC\) is isosceles.

- Compare the two right triangles formed by the altitude. - Which hypotenuses are congruent, and which leg is shared? - What does triangle congruence imply about the two base segments?

1. Triangles \(ADC\) and \(BDC\) are right triangles. Their hypotenuses are congruent because \(AC = BC\), and they share leg \(CD\). 2. Therefore, \(\triangle ADC \cong \triangle BDC\) by the Hypotenuse-Leg theorem. 3. Corresponding base segments are congruent, so the altitude bisects \(\overline{AB}\). 4. Thus, \(AD = DB = \frac{14\,\text{cm}}{2} = 7\,\text{cm}\).

\(AD = 7\,\text{cm}\) and \(DB = 7\,\text{cm}\).

- Identify two triangles that contain \(CM\) and \(DM\). - Which angle pairs come from vertical angles and parallel lines? - What follows about corresponding sides after proving the triangles congruent?

1. Compare \(\triangle BCM\) and \(\triangle FDM\). 2. Since \(M\) is the midpoint of \(\overline{CD}\), \(CM = DM\). 3. Angles \(\angle BMC\) and \(\angle FMD\) are vertical angles, so they are congruent. 4. Because \(BC \parallel AF\), angles \(\angle BCM\) and \(\angle FDM\) are alternate interior angles, so they are congruent. 5. Therefore, \(\triangle BCM \cong \triangle FDM\) by ASA. 6. Corresponding sides of congruent triangles are congruent, so \(BC = DF\).

\(BC = DF\) because \(\triangle BCM \cong \triangle FDM\) by ASA.

- Use the definition of midpoint on both intersecting segments. - Identify the vertical angles at \(Z\). - Apply corresponding parts after proving congruence. - Match the vertices in the correct order.

1. Since \(Z\) is the midpoint of both segments, \(AZ = DZ\) and \(BZ = CZ\). 2. Angles \(\angle AZB\) and \(\angle DZC\) are vertical angles, so they are congruent. 3. Therefore, \(\triangle ABZ \cong \triangle DCZ\) by SAS. 4. Corresponding parts of congruent triangles give \(\angle ZDC = \angle ZAB = 35^\circ\) and \(CD = AB = 4\,\text{cm}\).

a) \(\alpha = 35^\circ\) b) \(CD = 4\,\text{cm}\) c) SAS

- Look for pairs of triangles with equal halves of \(BC\) and equal halves of \(AD\). - What kind of angle pairs are formed at \(M\)? - After proving triangles congruent, match their corresponding sides.

1. In \(\triangle ABM\) and \(\triangle DCM\), \(BM = MC\), \(AM = MD\), and \(\angle AMB \cong \angle DMC\) because they are vertical angles. 2. Therefore, \(\triangle ABM \cong \triangle DCM\) by SAS, so \(CD = AB = 9\,\text{cm}\). 3. In \(\triangle ACM\) and \(\triangle DBM\), \(CM = MB\), \(AM = MD\), and \(\angle AMC \cong \angle DMB\) because they are vertical angles. 4. Therefore, \(\triangle ACM \cong \triangle DBM\) by SAS, so \(BD = AC = 7\,\text{cm}\).

\(BD = 7\,\text{cm}\) and \(CD = 9\,\text{cm}\).

- Use the side and angle properties of an equilateral triangle. - Subtract equal segments from equal side lengths. - Compare the three corner triangles. - Which congruence theorem applies?

1. Since \(ABC\) is equilateral, \(AB = BC = CA\), and each angle measures \(60^{\circ}\). 2. Because \(AD = BE = CF\) and the full side lengths are equal, the remaining segments are also equal: \(BD = CE = AF\). 3. In triangles \(ADF\), \(BED\), and \(CFE\), two corresponding sides and the included \(60^{\circ}\) angle are congruent. 4. Therefore, the three corner triangles are congruent by SAS. 5. Corresponding third sides are congruent, so \(DF = DE = EF\). Thus \(DEF\) is equilateral.

The three corner triangles are congruent by SAS, so their corresponding third sides satisfy \(DF = DE = EF\). Therefore, \(DEF\) is equilateral.

- Use one pair of congruent opposite sides. - Find alternate interior angles made by each diagonal. - Identify the congruence criterion from two angles and their included side.

1. Opposite sides of a parallelogram are congruent, so \(AB = CD\). 2. Since \(AB \parallel CD\), transversal \(\overline{AC}\) gives \(\angle BAS = \angle DCS\), and transversal \(\overline{BD}\) gives \(\angle ABS = \angle CDS\). 3. The equal side lies between the two pairs of equal angles. 4. Therefore, \(\triangle ABS \cong \triangle CDS\) by ASA.

\(\triangle ABS \cong \triangle CDS\) by ASA, using opposite sides of a parallelogram and alternate interior angles formed by parallel lines.

- Find two triangles containing \(\overline{BG}\) and \(\overline{EC}\). - Use the congruent sides and right angles of the squares. - Consider a \(90^\circ\) rotation about \(A\). - Identify two sides and the included angle for SAS.

1. Compare \(\triangle ABG\) and \(\triangle AEC\). 2. Because \(ABDE\) is a square, \(AB = AE\). 3. Because \(ACFG\) is a square, \(AG = AC\). 4. A \(90^\circ\) rotation about \(A\) maps ray \(\overrightarrow{AE}\) to ray \(\overrightarrow{AB}\) and ray \(\overrightarrow{AC}\) to ray \(\overrightarrow{AG}\). Rotations preserve angle measure, so \(\angle EAC = \angle BAG\). 5. Therefore, \(\triangle ABG \cong \triangle AEC\) by SAS. 6. Corresponding sides of congruent triangles are congruent, so \(BG = EC\).

The square sides give \(AB = AE\) and \(AG = AC\). A \(90^\circ\) rotation about \(A\) shows \(\angle BAG = \angle EAC\). Thus, \(\triangle ABG \cong \triangle AEC\) by SAS, and \(BG = EC\).

- Use the midpoint and extension conditions to identify two pairs of congruent segments. - Identify the vertical angles at \(M\). - Apply the triangle inequality to side \(\overline{AD}\) of \(\triangle ACD\). - Express \(AD\) in terms of the median length.

1. Since \(M\) is the midpoint of \(\overline{BC}\), \(BM = MC\). The construction gives \(AM = MD\). 2. Angles \(\angle AMB\) and \(\angle DMC\) are vertical angles, so they are congruent. 3. Therefore, \(\triangle ABM \cong \triangle DCM\) by SAS. Corresponding sides give \(CD = AB = c\). 4. In \(\triangle ACD\), the triangle inequality gives \(AD < AC + CD\). 5. Since \(AD = AM + MD = 2s_a\), \(AC = b\), and \(CD = c\), substitution gives \(2s_a < b + c\). 6. Dividing by \(2\) gives \(s_a < \frac{b + c}{2}\).

a) \(\triangle ABM \cong \triangle DCM\) by SAS, so \(CD = AB = c\). b) From \(AD < AC + CD\), substitute \(AD = 2s_a\), \(AC = b\), and \(CD = c\). Then \(2s_a < b + c\), so \(s_a < \frac{b + c}{2}\).

- Which sides and angles are congruent in an equilateral triangle? - Look for two triangles containing the given congruent segments \(AM\) and \(BN\). - After proving those triangles congruent, express two angles of \(\triangle APC\) using one variable. - Use the triangle angle sum.

1. Compare \(\triangle ABN\) and \(\triangle CAM\). 2. Since \(\triangle ABC\) is equilateral, \(AB = CA\) and \(m\angle ABN = m\angle CAM = 60^\circ\). Also, \(BN = AM\) is given. 3. Therefore, \(\triangle ABN \cong \triangle CAM\) by SAS. 4. Corresponding angles give \(\angle BAN \cong \angle ACM\). Let their common measure be \(x^\circ\). 5. In \(\triangle APC\), \(m\angle PAC = 60^\circ - x^\circ\) and \(m\angle PCA = x^\circ\). 6. Thus, \(m\angle APC = 180^\circ - (60^\circ - x^\circ) - x^\circ = 120^\circ\).

\(m\angle APC = 120^\circ\)

- Begin with the base angles of \(\triangle ADE\). - Relate those angles to the adjacent angles on line \(BC\). - Apply a congruence criterion to the two outer triangles.

1. Since \(AD = AE\), \(\triangle ADE\) is isosceles, so \(\angle ADE = \angle AED\). 2. Because \(B,D,E,C\) are collinear, \(\angle ADB\) and \(\angle ADE\) are supplementary, and \(\angle AEC\) and \(\angle AED\) are supplementary. Therefore, \(\angle ADB = \angle AEC\). 3. In \(\triangle ADB\) and \(\triangle AEC\), \(AD = AE\), \(BD = CE\), and the included angles \(\angle ADB\) and \(\angle AEC\) are congruent. 4. Thus, \(\triangle ADB \cong \triangle AEC\) by SAS. 5. Corresponding sides of congruent triangles are congruent, so \(AB = AC\). Therefore, \(\triangle ABC\) is isosceles.

The base angles of isosceles \(\triangle ADE\) are congruent, so their supplementary angles \(\angle ADB\) and \(\angle AEC\) are congruent. Then \(\triangle ADB \cong \triangle AEC\) by SAS, which gives \(AB = AC\). Therefore, \(\triangle ABC\) is isosceles.

- Compare the two triangles formed by the intersecting segments. - What do the midpoint markings tell you about side lengths? - Which angles at an intersection are always congruent? - Which corresponding angles can be used to prove the lines parallel?

1. Since \(K\) is the midpoint of both segments, \(AK = KC\) and \(BK = KD\). 2. Angles \(\angle AKB\) and \(\angle CKD\) are vertical angles, so they are congruent. 3. Therefore, \(\triangle AKB \cong \triangle CKD\) by SAS. 4. Corresponding parts of congruent triangles give \(\angle KAB \cong \angle KCD\). 5. These are alternate interior angles formed by transversal \(AC\). By the converse of the alternate interior angles theorem, \(AB \parallel CD\).

The triangles are congruent by SAS, so \(\angle KAB \cong \angle KCD\). Since these are alternate interior angles, \(AB \parallel CD\).

- Compare the two triangles that use base \(\overline{PQ}\). - Use the base angles of the larger isosceles triangle. - Relate the corresponding angles to \(\triangle PQX\).

1. Since \(PR = QR\), the base angles satisfy \(\angle QPR = \angle PQR\). 2. In \(\triangle PQS\) and \(\triangle QPT\), \(PQ = QP\), \(PS = QT\), and \(\angle QPS = \angle PQT\). 3. Therefore, \(\triangle PQS \cong \triangle QPT\) by SAS. 4. Corresponding angles are congruent, so \(\angle PQS = \angle QPT\). 5. Because \(X\) lies on \(\overline{QS}\) and \(\overline{PT}\), these are the base angles \(\angle PQX\) and \(\angle QPX\) of \(\triangle PQX\). 6. Thus, \(\angle PQX = \angle QPX\), so \(\triangle PQX\) is isosceles.

\(\triangle PQS \cong \triangle QPT\) by SAS, so \(\angle PQX = \angle QPX\). Therefore, \(\triangle PQX\) is isosceles.

- Subtract the given equal segments from congruent sides of the outer parallelogram. - Compare the small triangles at opposite vertices using SAS. - Which parallelogram test uses two pairs of congruent opposite sides?

1. Opposite sides of parallelogram \(KPHT\) are congruent, so \(KP = HT\) and \(PH = TK\). 2. Since \(PB = TD\), subtraction gives \(BK = KP - PB = HT - TD = HD\). 3. Since \(CH = AK\), subtraction gives \(PC = PH - CH = TK - AK = AT\). 4. Opposite angles of the outer parallelogram are congruent: \(\angle K \cong \angle H\) and \(\angle P \cong \angle T\). 5. Therefore, \(\triangle ABK \cong \triangle CDH\) by SAS, so \(AB = CD\). Also, \(\triangle BCP \cong \triangle DAT\) by SAS, so \(BC = DA\). 6. Both pairs of opposite sides of \(ABCD\) are congruent. Therefore, \(ABCD\) is a parallelogram.

SAS proves the opposite corner triangles congruent, giving \(AB = CD\) and \(BC = DA\). Since both pairs of opposite sides are congruent, \(ABCD\) is a parallelogram.

- Use the diagonal intersection \(O\) of the inner parallelogram. - Show that \(BO = DO\) by adding equal collinear segments. - Compare \(\triangle AOB\) and \(\triangle COD\) using SAS.

1. Let \(O\) be the intersection of the diagonals of parallelogram \(AKCF\). Then \(AO = OC\) and \(KO = OF\). 2. Since \(B\), \(K\), \(O\), \(F\), and \(D\) are collinear, \(BO = BK + KO\) and \(DO = DF + FO\). 3. Using \(BK = DF\) and \(KO = FO\), it follows that \(BO = DO\). 4. Angles \(\angle AOB\) and \(\angle COD\) are vertical angles, so they are congruent. 5. Therefore, \(\triangle AOB \cong \triangle COD\) by SAS. 6. Corresponding parts give \(AB = CD\) and \(\angle ABO \cong \angle CDO\). These are alternate interior angles formed by transversal \(BD\), so the converse of the alternate interior angles theorem gives \(AB \parallel CD\). 7. One pair of opposite sides is both congruent and parallel, so \(ABCD\) is a parallelogram.

Parallelogram \(AKCF\) gives \(AO = OC\) and \(KO = OF\). Together with \(BK = FD\), this gives \(BO = DO\). SAS then proves \(\triangle AOB \cong \triangle COD\), so \(AB = CD\) and \(\angle ABO \cong \angle CDO\). By the converse of the alternate interior angles theorem, \(AB \parallel CD\). Therefore, \(ABCD\) is a parallelogram.

- Draw perpendiculars from the endpoints of the shorter base to the longer base. - Compare the right triangles containing the congruent diagonals. - Use the resulting angle congruence to compare two larger triangles.

1. Let \(ABCD\) be a trapezoid with \(AB \parallel CD\) and \(AC = BD\). 2. Draw altitudes \(DE\) and \(CF\) to \(AB\). Since \(AB \parallel CD\), the perpendicular distance between the parallel lines is constant, so \(DE = CF\). 3. Right triangles \(BDE\) and \(ACF\) have congruent hypotenuses \(BD = AC\) and congruent legs \(DE = CF\). They are congruent by HL. 4. Therefore, \(\angle DBA = \angle CAB\). 5. In triangles \(ABD\) and \(BAC\), \(AB\) is common, \(BD = AC\), and the included angles \(\angle DBA\) and \(\angle CAB\) are congruent. Thus, the triangles are congruent by SAS. 6. Corresponding legs satisfy \(AD = BC\), so \(ABCD\) is an isosceles trapezoid.

The altitude construction gives \(\triangle BDE \cong \triangle ACF\) by HL, which leads to \(\triangle ABD \cong \triangle BAC\) by SAS. Hence \(AD = BC\), so the trapezoid is isosceles.

- Use the right angles of both squares. - Compare the hypotenuses of the corner triangles. - Relate adjacent acute angles using complementary angles. - Apply the same argument around the figure.

1. Each corner triangle is a right triangle because every corner of the outer square measures \(90^\circ\). 2. The hypotenuses \(SP\), \(PQ\), \(QR\), and \(RS\) are sides of the inner square, so they are congruent. 3. Let \(\alpha = \angle APS\). Since \(A,P,B\) are collinear and \(\angle SPQ = 90^\circ\), \(\angle QPB = 90^\circ - \alpha\). 4. In the right triangles, \(\angle ASP = 90^\circ - \alpha\) and \(\angle BQP = \alpha\). Thus, \(\angle APS = \angle BQP\), \(\angle ASP = \angle QPB\), and the included sides \(SP\) and \(QP\) are congruent. 5. Therefore, \(\triangle APS \cong \triangle BQP\) by ASA. Repeating the same argument at \(Q\), \(R\), and \(S\) proves that all four corner triangles are congruent.

All four corner triangles are congruent by ASA: their hypotenuses are equal sides of the inner square, and the adjacent acute angles match cyclically.

- First relate the given exterior angles to the base angles of \(\triangle APQ\). - Use the converse of the isosceles triangle theorem. - Then compare \(\triangle ABP\) and \(\triangle ACQ\).

1. Since \(B,P,Q,C\) are collinear, \(\angle APB\) and \(\angle APQ\) are supplementary. Likewise, \(\angle AQC\) and \(\angle AQP\) are supplementary. 2. The given exterior angles are congruent, so their supplements are congruent: \(\angle APQ = \angle AQP\). 3. Therefore, \(\triangle APQ\) is isosceles and \(AP = AQ\). 4. In \(\triangle ABP\) and \(\triangle ACQ\), \(AP = AQ\), \(BP = CQ\), and \(\angle APB = \angle AQC\). 5. Thus, \(\triangle ABP \cong \triangle ACQ\) by SAS. 6. Corresponding sides are congruent, so \(AB = AC\).

The equal exterior angles imply \(\angle APQ = \angle AQP\), so \(AP = AQ\). Then \(\triangle ABP \cong \triangle ACQ\) by SAS, giving \(AB = AC\).

- First prove \(DE = DF\) by subtracting equal segments. - Use SAS on \(\triangle ADF\) and \(\triangle CDE\). - Compare the remaining parts of the equal base angles of \(\triangle ADC\).

1. Since \(AD = CD\) and \(AE = CF\), subtract equal lengths to obtain \(DE = DF\). 2. In \(\triangle ADF\) and \(\triangle CDE\), \(AD = CD\), \(DF = DE\), and the angle at \(D\) is common. 3. Therefore, \(\triangle ADF \cong \triangle CDE\) by SAS, so \(\angle DAF = \angle DCE\). 4. The base angles of isosceles \(\triangle ADC\) satisfy \(\angle DAC = \angle DCA\). 5. Since \(B\) lies on \(\overline{AF}\) and \(\overline{CE}\), subtracting the equal smaller angles gives \(\angle BAC = \angle BCA\). 6. By the converse of the isosceles triangle theorem, \(AB = CB\). Thus, \(\triangle ABC\) is isosceles.

\(\triangle ADF \cong \triangle CDE\) by SAS, so corresponding angle subtraction gives \(\angle BAC = \angle BCA\). Therefore, \(AB = CB\), and \(\triangle ABC\) is isosceles.