Free math worksheets

- Focus on the two lines that cross at \(M\). - What name is given to opposite angles formed by two intersecting lines?

1. Lines \(AC\) and \(BD\) intersect at \(M\). 2. Angles \(\angle AMB\) and \(\angle CMD\) are opposite angles formed by the intersecting lines, so they are vertical angles. 3. Vertical angles are congruent, which supplies the missing reason.

Vertical angles are congruent.

- 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) \(PD = QB\).

- What does being the midpoint of a segment tell you about a half-turn centered there? - Track both endpoints of \(\overline{AB}\). - Recall which properties of a segment and its supporting line a rotation preserves.

1. Since \(K\) is the midpoint of \(\overline{AC}\), a \(180^\circ\) rotation about \(K\) maps \(A\) to \(C\). 2. Since \(K\) is also the midpoint of \(\overline{BD}\), the same rotation maps \(B\) to \(D\). 3. Therefore, segment \(\overline{AB}\) maps to segment \(\overline{CD}\). 4. A rotation preserves length, so \(AB=CD\). 5. A \(180^\circ\) rotation maps a line not passing through its center to a parallel line, so \(AB\parallel CD\).

The half-turn about \(K\) maps \(A\to C\) and \(B\to D\), so \(\overline{AB}\) maps to \(\overline{CD}\). Therefore, \(AB=CD\) and \(AB\parallel CD\).

- Identify the two right angles created by the altitudes. - Compare the angles at vertex \(B\) in the two triangles. - Check whether the given congruent side lies between the two known angles.

1. Because \(AD\) and \(CE\) are altitudes, \(\angle ADB\) and \(\angle CEB\) are right angles, so they are congruent. 2. Since \(D\) lies on \(BC\) and \(E\) lies on \(BA\), angles \(\angle ABD\) and \(\angle CBE\) are the same angle formed by lines \(BA\) and \(BC\). 3. The nonincluded sides satisfy \(AB=BC\) because \(ABC\) is isosceles. 4. Therefore, \(\triangle ABD\cong\triangle CBE\) by AAS. 5. By CPCTC, \(AD=CE\).

a) \(\triangle ABD\cong\triangle CBE\) by AAS. b) CPCTC gives \(AD=CE\).

- 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.

- 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, \(AD = BD\).

- 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.

- Identify the hypotenuse and a pair of congruent legs. - Which theorem proves two right triangles congruent from that information?

1. Both triangles are right triangles: \(\angle ABD = \angle ACD = 90^{\circ}\). 2. They share hypotenuse \(\overline{AD}\). 3. One pair of corresponding legs is congruent: \(\overline{AB} = \overline{CD}\). 4. Therefore, \(\triangle ABD \cong \triangle DCA\) by the Hypotenuse–Leg theorem.

\(\triangle ABD \cong \triangle DCA\) by HL.

- 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 \(A\).

- Count the equal side pairs available after using the shared diagonal. - Match the vertex correspondence in \(\triangle ABC\cong\triangle ADC\). - The final angle must be the angle at \(A\) on the other side of \(AC\).

1. The diagram gives \(AB=AD\) and \(CB=CD\). 2. Segment \(AC\) is shared by both triangles. 3. Thus, all three corresponding side pairs are congruent, so \(\triangle ABC\cong\triangle ADC\) by SSS. 4. Corresponding angles then give \(\angle BAC\cong\angle DAC\), so \(AC\) bisects \(\angle BAD\).

3. SSS 4. \(\angle DAC\)

- 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.

- Compare the two sides that meet at \(A\) in each square. - A square fixes both a right-angle turn and an equal distance from its vertex. - Track the two endpoints of \(\overline{EC}\) under the same rotation.

1. In square \(ABDE\), ray \(\overrightarrow{AE}\) turns \(90^\circ\) counterclockwise to ray \(\overrightarrow{AB}\), and \(AE=AB\). Therefore, a \(90^\circ\) counterclockwise rotation about \(A\) maps \(E\) to \(B\). 2. In square \(ACFG\), the same \(90^\circ\) counterclockwise rotation about \(A\) maps \(C\) to \(G\). 3. Therefore, the rotation maps segment \(\overline{EC}\) to segment \(\overline{BG}\). 4. Rotations preserve distance, so \(EC=BG\).

A \(90^\circ\) counterclockwise rotation about \(A\) maps \(E\to B\) and \(C\to G\). Thus, \(\overline{EC}\) maps to \(\overline{BG}\), and distance preservation gives \(EC=BG\).

- Compare the two triangles on opposite sides of \(\overline{BD}\). - Identify the shared side and the two angles created by the angle bisector. - After proving the triangles congruent, focus on the distances from \(B\) to \(A\) and \(C\).

1. In triangles \(ADB\) and \(CDB\), \(AD=CD\) is given and \(DB=DB\) by the reflexive property. 2. Because \(\overline{BD}\) bisects \(\angle ADC\), \(\angle ADB\cong\angle CDB\). 3. Therefore, \(\triangle ADB\cong\triangle CDB\) by SAS. 4. Corresponding sides are congruent, so \(AB=CB\). 5. Since \(B\) is equidistant from the endpoints of \(\overline{AC}\), the converse of the Perpendicular Bisector Theorem shows that \(B\) lies on the perpendicular bisector of \(\overline{AC}\).

a) \(\triangle ADB\cong\triangle CDB\) by SAS. b) CPCTC gives \(AB=CB\). Therefore, by the converse of the Perpendicular Bisector Theorem, \(B\) lies on the perpendicular bisector of \(\overline{AC}\).

- Use the midpoint, the isosceles-triangle side equality, and the shared segment for part a). - After congruence, identify the corresponding angles at \(D\). - What must two congruent angles that form a linear pair measure? - Connect the resulting line to the definition of a perpendicular bisector.

1. Since \(D\) is the midpoint of \(\overline{AC}\), \(AD=DC\). Also, \(AE=CE\) is given and \(ED=ED\) by the reflexive property. 2. Therefore, \(\triangle AED\cong\triangle CED\) by SSS. 3. CPCTC gives \(\angle ADE\cong\angle EDC\). 4. Because \(A\), \(D\), and \(C\) are collinear, these congruent angles form a linear pair. Therefore, each is a right angle, so \(ED\perp AC\). 5. Line \(ED\) passes through the midpoint of \(\overline{AC}\) and is perpendicular to \(AC\), so it is the perpendicular bisector of \(\overline{AC}\). 6. Since \(B\) lies on line \(ED\), the Perpendicular Bisector Theorem gives \(AB=CB\).

a) \(\triangle AED\cong\triangle CED\) by SSS. b) CPCTC gives \(\angle ADE\cong\angle EDC\); as a linear pair, they are both right angles, so \(ED\perp AC\). c) \(ED\) is the perpendicular bisector of \(AC\), so \(AB=CB\).

- Start with what the diagonals of the original parallelogram tell you about \(S\). - Compare the remaining distances from \(P\) and \(Q\) to \(S\). - A half-turn pairs points on opposite rays when they are the same distance from the center. - Connect those point mappings to the diagonals of the new quadrilateral.

1. The diagonals of parallelogram \(ABCD\) bisect each other, so \(S\) is the midpoint of both \(\overline{AC}\) and \(\overline{BD}\). Therefore, a \(180^\circ\) rotation about \(S\) maps \(A\leftrightarrow C\) and \(B\leftrightarrow D\). 2. Because \(BS=DS\) and \(BP=DQ\), subtracting equal lengths gives \(SP=SQ\). 3. Points \(P\) and \(Q\) lie on opposite rays of line \(BD\) from \(S\), so the half-turn maps \(P\leftrightarrow Q\). 4. Thus, the rotation maps \(\overline{AP}\) to \(\overline{CQ}\) and \(\overline{AQ}\) to \(\overline{CP}\). 5. The half-turn therefore maps quadrilateral \(APCQ\) onto itself with opposite vertices paired. Its diagonals \(AC\) and \(PQ\) have the same midpoint \(S\), so they bisect each other. 6. A quadrilateral whose diagonals bisect each other is a parallelogram. Therefore, \(APCQ\) is a parallelogram.

The \(180^\circ\) rotation about \(S\) maps \(A\leftrightarrow C\). From \(BS=DS\) and \(BP=DQ\), it follows that \(SP=SQ\), so the same half-turn maps \(P\leftrightarrow Q\). Hence, \(AC\) and \(PQ\) bisect each other at \(S\), and \(APCQ\) 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 \(BD=8\,\text{cm}\).

- The proof is establishing the parallelogram test, so you may not invoke that test as a shortcut. - Use the given parallel sides to obtain an angle pair involving diagonal \(AC\). - After triangle congruence, look for a corresponding angle pair that can prove the other sides parallel.

1. Draw diagonal \(\overline{AC}\). 2. Since \(AB\parallel CD\), \(\angle BAC\cong\angle ACD\) by the alternate interior angles theorem. 3. In \(\triangle ABC\) and \(\triangle CDA\), \(AB=CD\) is given, \(AC=CA\) by the reflexive property, and the included angles are congruent. 4. Therefore, \(\triangle ABC\cong\triangle CDA\) by SAS. 5. CPCTC gives \(\angle BCA\cong\angle CAD\). 6. These are alternate interior angles for lines \(BC\) and \(AD\) cut by transversal \(AC\). By the converse of the alternate interior angles theorem, \(BC\parallel AD\). 7. Now both pairs of opposite sides are parallel, so \(ABCD\) is a parallelogram.

Diagonal \(AC\) creates triangles \(ABC\) and \(CDA\). The given parallel sides supply one included-angle pair, so SAS proves the triangles congruent. CPCTC supplies a second alternate-interior-angle pair, which proves \(BC\parallel AD\). Therefore, both pairs of opposite sides are parallel and \(ABCD\) is a parallelogram.

- 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.

- 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 information for one side pair. - The altitude and the location of \(B\) determine the included angles at \(D\). - Identify the side shared by the two triangles.

1. Since \(D\) is the midpoint of \(\overline{AC}\), \(AD=CD\). 2. Because \(ED\perp AC\) and \(B\) lies on \(ED\), \(\angle ADB\) and \(\angle CDB\) are right angles, so they are congruent. 3. The triangles share side \(BD\), so \(BD=BD\). 4. Therefore, \(\triangle ABD\cong\triangle CBD\) by SAS. 5. By CPCTC, \(AB=CB\). Thus, \(\triangle ABC\) is isosceles.

a) \(\triangle ABD\cong\triangle CBD\) by SAS. b) CPCTC gives \(AB=CB\), so \(\triangle ABC\) is isosceles.

- Identify the reflection symmetry of the outer isosceles triangle. - Track a point on one congruent side by its distance from the corresponding base vertex. - What happens to the intersection of two lines when the reflection swaps those two lines?

1. The reflection across the symmetry axis of isosceles \(\triangle ADC\)—the line through \(D\) and the midpoint of \(AC\)—maps \(A\leftrightarrow C\) and side \(AD\) onto side \(CD\). 2. Reflection preserves distance. The image of \(E\) must lie on \(CD\) at the same distance from \(C\) that \(E\) lies from \(A\). Since \(AE=CF\), the image of \(E\) is \(F\). 3. Therefore, line \(CE\) reflects to line \(AF\), and line \(AF\) reflects to line \(CE\). 4. Their intersection \(B\) must map to itself, so \(B\) lies on the mirror line. 5. Since the reflection maps \(A\) to \(C\) and fixes \(B\), it maps \(\overline{BA}\) to \(\overline{BC}\). Reflections preserve distance, so \(BA=BC\). 6. Therefore, \(\triangle ABC\) is isosceles.

Reflect across the symmetry axis through \(D\) and the midpoint of \(AC\). The reflection maps \(A\leftrightarrow C\) and, because \(AE=CF\), maps \(E\leftrightarrow F\). Thus, lines \(AF\) and \(CE\) are images of each other, so their intersection \(B\) is fixed. Hence, \(BA=BC\), and \(\triangle ABC\) is isosceles.

- Begin with the half-turn symmetry of the given parallelogram. - Compare the distances from \(B\) and \(D\) to the rotation center by adding equal collinear pieces. - A half-turn pairs points on opposite rays when their distances from the center agree.

1. The diagonals of parallelogram \(AKCF\) bisect each other at \(O\). Therefore, a \(180^\circ\) rotation about \(O\) maps \(A\leftrightarrow C\) and \(K\leftrightarrow F\). 2. Because \(KO=FO\) and \(BK=FD\), segment addition gives \(BO=DO\). 3. Points \(B\) and \(D\) lie on opposite rays of line \(KF\) from \(O\), so the half-turn maps \(B\leftrightarrow D\). 4. Thus, the rotation maps \(\overline{AB}\) to \(\overline{CD}\) and \(\overline{BC}\) to \(\overline{DA}\). 5. In particular, \(O\) is the midpoint of both \(\overline{AC}\) and \(\overline{BD}\). Therefore, the diagonals of \(ABCD\) bisect each other. 6. Hence, \(ABCD\) is a parallelogram.

The \(180^\circ\) rotation about \(O\) maps \(A\leftrightarrow C\), \(K\leftrightarrow F\), and, using \(BK=FD\), \(B\leftrightarrow D\). Thus, \(O\) is the midpoint of both diagonals \(AC\) and \(BD\), so \(ABCD\) is a parallelogram.

- Use the midpoint to identify the side included between the relevant angles. - Find one angle pair from the intersection at \(M\). - Find another angle pair from the parallel bases.

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

a) \(\triangle BCM\cong\triangle FDM\) by ASA. b) CPCTC gives \(BC=DF\).

- Use the definition of midpoint on both intersecting segments. - Identify the vertical angles at \(Z\). - After proving congruence, match the requested angle and side to their corresponding parts.

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. By CPCTC, \(\angle ZDC=\angle ZAB=35^\circ\) and \(CD=AB=4\,\text{cm}\).

a) \(\triangle ABZ\cong\triangle DCZ\) by SAS. b) \(\alpha=35^\circ\) c) \(CD=4\,\text{cm}\)

- 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.

- First identify which statements are raw givens or reflexive facts and which require triangle congruence. - A CPCTC statement cannot be justified until the triangle-congruence statement has been established. - Each parallel-line conclusion needs the matching angle congruence before it. - The final quadrilateral classification needs both parallel-side conclusions.

1. Start with E, the given side congruences. 2. Use D, the reflexive fact \(AC=CA\). 3. From E and D, the three corresponding side pairs are congruent, so C follows by SSS. 4. From C, CPCTC gives both A and F. These two statements may appear in either order. 5. From A, the converse of the alternate interior angles theorem gives B. 6. From F, the converse of the alternate interior angles theorem gives G. 7. Once B and G are established, H follows from the definition of a parallelogram. 8. One valid order is E, D, C, A, B, F, G, H.

One valid order is E, D, C, A, B, F, G, H. D: reflexive property. C: SSS. A: CPCTC. B: converse of the alternate interior angles theorem. F: CPCTC. G: converse of the alternate interior angles theorem. H: both pairs of opposite sides are parallel, so \(ABCD\) is a parallelogram. A and F may be interchanged, and the two resulting parallel-line deductions may be interleaved as long as each comes after its required CPCTC statement.

- The conclusion requires both a midpoint and a perpendicular line. - What point on \(\overline{AB}\) would give you a second pair of equal sides? - After proving those triangles congruent, look at the adjacent angles at the midpoint.

1. Construct midpoint \(M\) of \(\overline{AB}\) and draw \(\overline{PM}\). 2. Then \(PA=PB\) is given, \(AM=BM\) by the midpoint construction, and \(PM=PM\) by the reflexive property. 3. Therefore, \(\triangle PAM\cong\triangle PBM\) by SSS. 4. Corresponding angles \(\angle PMA\) and \(\angle PMB\) are congruent. They form a linear pair, so each is \(90^\circ\). 5. Thus, \(PM\perp AB\), and because \(M\) is the midpoint of \(AB\), line \(PM\) is the perpendicular bisector. Therefore, \(P\) lies on it.

Construct the midpoint \(M\) of \(\overline{AB}\) and draw \(\overline{PM}\). This creates two triangles that are congruent by SSS and leads to \(PM\perp AB\).

- 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 \(s_a < \frac{b + c}{2}\).

- Start with the rotational symmetry of an equilateral triangle. - Track a point on one side by its distance from a vertex. - Compare the directions of the two intersecting segments after the rotation.

1. A \(120^\circ\) counterclockwise rotation about the center of equilateral triangle \(ABC\) maps \(A\to B\), \(B\to C\), and \(C\to A\). 2. Point \(M\) lies on \(AB\) at distance \(AM\) from \(A\). Its image lies on \(BC\) at the same distance from \(B\). Since \(AM=BN\), the image of \(M\) is \(N\). 3. Therefore, the directed line from \(M\) toward \(C\) rotates onto the directed line from \(N\) toward \(A\). 4. Ray \(\overrightarrow{PC}\) has the same direction as \(\overrightarrow{MC}\), and ray \(\overrightarrow{PA}\) has the same direction as \(\overrightarrow{NA}\). The rotation between those directions is \(120^\circ\). 5. Hence, \(m\angle APC=120^\circ\).

The \(120^\circ\) counterclockwise rotation maps \(A\to B\to C\to A\) and, because \(AM=BN\), maps \(M\to N\). It maps the direction of \(MC\) to the direction of \(NA\), so \(m\angle APC=120^\circ\).

- 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.

- Pair opposite sides of the outer parallelogram before subtracting the given equal pieces. - Match each small corner triangle with the triangle at the opposite vertex. - Use opposite angles of the outer parallelogram as the included-angle pairs. - After CPCTC, choose the parallelogram test supported by the two resulting side equalities.

1. Opposite sides of parallelogram \(KPHT\) are congruent, so \(KP=HT\) and \(PH=TK\). 2. Since \(PB=TD\), subtracting equal lengths from \(KP=HT\) gives \(BK=HD\). 3. Since \(CH=AK\), subtracting equal lengths from \(PH=TK\) gives \(PC=AT\). 4. Opposite angles of the outer parallelogram are congruent, so \(\angle AKB\cong\angle CHD\). Together with \(AK=CH\) and \(BK=HD\), SAS gives \(\triangle ABK\cong\triangle CDH\). 5. Likewise, \(\angle BPC\cong\angle DTA\). Together with \(PB=TD\) and \(PC=AT\), SAS gives \(\triangle BCP\cong\triangle DAT\). 6. CPCTC gives \(AB=CD\) and \(BC=DA\). 7. A quadrilateral with both pairs of opposite sides congruent is a parallelogram. Therefore, \(ABCD\) is a parallelogram.

a) \(BK=HD\) and \(PC=AT\). b) \(\triangle ABK\cong\triangle CDH\) by SAS and \(\triangle BCP\cong\triangle DAT\) by SAS. c) CPCTC gives \(AB=CD\) and \(BC=DA\), so \(ABCD\) is a parallelogram.

- Try extending one base and constructing a segment equal in length to the other base. - What quadrilateral is formed when one pair of opposite sides is both parallel and congruent? - Use the auxiliary parallelogram and the congruent diagonals to create an angle equality for an SAS proof.

1. On ray \(AB\) beyond \(B\), construct point \(E\) so that \(BE = CD\). 2. Since \(BE \parallel CD\) and \(BE = CD\), quadrilateral \(BECD\) is a parallelogram. Therefore, \(EC \parallel BD\) and \(EC = BD\). 3. The given \(AC = BD\), together with \(EC = BD\), gives \(AC = EC\). Thus, \(\triangle ACE\) is isosceles and \(\angle CAE = \angle AEC\). 4. Because \(A\), \(B\), and \(E\) are collinear, \(\angle CAE = \angle CAB\). Because \(EA\) is collinear with \(BA\) and \(EC \parallel BD\), \(\angle AEC = \angle DBA\). Hence \(\angle CAB = \angle DBA\). 5. In \(\triangle CAB\) and \(\triangle DBA\), \(CA = DB\), \(AB = BA\), and the included angles \(\angle CAB\) and \(\angle DBA\) are congruent. Therefore, \(\triangle CAB \cong \triangle DBA\) by SAS. 6. Corresponding sides give \(CB = DA\). Therefore, \(ABCD\) is an isosceles trapezoid.

Extend ray \(AB\) beyond \(B\) and choose \(E\) so that \(BE=CD\). Then \(BECD\) is a parallelogram, leading to \(AC=EC\) and \(\angle CAB=\angle DBA\). Thus \(\triangle CAB\cong\triangle DBA\) by SAS, so \(AD=BC\).

- 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.

- Check the reason attached to every claimed angle relationship before checking the final criterion. - Compare the student's three-letter criterion with the valid triangle-congruence criteria in this course. - Look for a side relationship the student did not use.

1. The first error is the claim about the linear pair. Angles in a linear pair are supplementary, not automatically congruent. 2. The second error is the use of SSA. SSA is not a valid general triangle-congruence criterion. 3. The conclusion can still be proved. The given information gives \(AB = AC\) and, because \(D\) is the midpoint of \(BC\), \(BD = DC\). 4. The triangles also share \(\overline{AD}\), so \(AD = AD\). 5. Therefore, \(\triangle ABD\cong\triangle ACD\) by SSS.

The proof has two errors: a linear pair is supplementary, not necessarily congruent, and SSA is not a valid general congruence criterion. The proof is repaired by using \(AB = AC\), \(BD = DC\), and the shared side \(AD = AD\), so \(\triangle ABD\cong\triangle ACD\) by SSS.

- Use the marked side pairs to determine the vertex correspondence before reading the student's triangle order. - The first letters must correspond, the second letters must correspond, and the third letters must correspond. - A true final equality does not make an invalid correspondence statement acceptable.

1. The marked side \(AB\) corresponds to \(DE\), so \(A\leftrightarrow D\) and \(B\leftrightarrow E\). 2. The marked side \(AC\) corresponds to \(DF\), so \(C\leftrightarrow F\). 3. Therefore, the correct vertex order is \(\triangle ABC\cong\triangle DEF\), not \(\triangle DFE\). 4. With the correct correspondence, side \(BC\) corresponds to \(EF\), so CPCTC gives \(BC=EF\). 5. The student's numerical side conclusion used the same undirected segment \(FE=EF\), but the written congruence order was still logically incorrect.

The error is the vertex order in the congruence statement. The corrected statement is \(\triangle ABC\cong\triangle DEF\) by SAS, and CPCTC gives \(BC=EF\).

- List exactly which three facts each proposed criterion requires. - Do not forget the side shared by both triangles. - For part b), remove only the angle fact and check which proof still has all of its required information.

1. Route 1 uses \(AB=CD\), \(BC=AD\), and the shared side \(AC=CA\). Therefore, \(\triangle ABC\cong\triangle CDA\) by SSS. 2. Route 2 uses \(AB=CD\), \(BC=AD\), and the included-angle equality \(\angle ABC\cong\angle CDA\). Therefore, the same triangles are congruent by SAS. 3. If the angle equality is removed, Route 1 still works because all three side equalities remain available, including the reflexive side \(AC=CA\). 4. Therefore, the angle equality is sufficient for the SAS route but not necessary for the congruence conclusion; it is redundant once the three side pairs are known.

a) Route 1 is valid by SSS using \(AB=CD\), \(BC=AD\), and \(AC=CA\). Route 2 is valid by SAS using the two given side pairs and the included angle \(\angle ABC\cong\angle CDA\). b) Route 1 still works. c) The angle equality is redundant for proving these triangles congruent because SSS already suffices.

- 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 \(AB = AC\).