- 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.

- 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. \(\angle DAC\)

- Compare each marked side with the side carrying the same kind of tick in the other triangle. - Can a segment be a side of two triangles even when it is drawn only once? - Check that your side pairings agree with the vertex order of the congruence statement.

1. The first pair of matching marked sides gives \(AB=DC\). 2. The second pair gives \(AC=DB\). 3. The shared segment gives \(BC=CB\). 4. Therefore \(\triangle ABC\cong\triangle DCB\) by SSS.

\(AB=DC\), \(AC=DB\), and \(BC=CB\) (shared side). Thus \(\triangle ABC\cong\triangle DCB\) by SSS.

- The marked sides meet at the crossing point. - Which angles at an intersection always have equal measure? - Check whether that angle lies between the two known sides.

1. The marked side pairs give \(AM=CM\) and \(BM=DM\). 2. Since the straight lines cross at \(M\), \(\angle AMB=\angle CMD\) as vertical angles. 3. These are the included angles between the equal sides. 4. Therefore \(\triangle AMB\cong\triangle CMD\) by SAS.

\(AM=CM\), \(BM=DM\), and \(\angle AMB=\angle CMD\) (vertical angles). Therefore \(\triangle AMB\cong\triangle CMD\) by SAS.

- Locate the side between the two named angles in each triangle. - State the criterion after listing its matching parts. - Read the final side pair from the order of the congruence statement.

1. The two given angle pairs correspond: \(\angle P=\angle S\) and \(\angle Q=\angle T\). 2. The given equal sides \(PQ=ST\) lie between the two known angles. 3. Thus \(\triangle PQR\cong\triangle STU\) by ASA. 4. The matching side pair \(PR\) and \(SU\) must have equal lengths because corresponding parts of congruent triangles are equal.

\(\angle P=\angle S\), \(\angle Q=\angle T\), and the included sides \(PQ=ST\). Thus \(\triangle PQR\cong\triangle STU\) by ASA, so \(PR=SU\).

- Check the vertex order so corresponding angles are paired correctly. - The position of a side, not just its length, decides whether ASA or AAS fits. - For the proof, explain how the given side connects to the two equal-angle pairs.

1. The given angle pairs match: \(\angle A=\angle D=45^\circ\) and \(\angle B=\angle E=70^\circ\). 2. The equal side \(BC=EF\) is not between those two angles in either triangle. 3. Therefore \(\triangle ABC\cong\triangle DEF\) by AAS.

\(\angle A=\angle D\), \(\angle B=\angle E\), and \(BC=EF\) is not included between these angles. Therefore \(\triangle ABC\cong\triangle DEF\) by AAS.

- 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 \(\angle ABC = \angle ADC\).

- What equal sides and right angles come from the rectangle? - Where do the two given short segments lie? - After finding a congruence criterion, match the remaining sides.

1. Opposite sides of a rectangle are equal, so \(AD=CB\). 2. The included angles \(\angle DAP\) and \(\angle BCQ\) are both right angles. 3. Together with \(AP=CQ\), these facts prove \(\triangle APD\cong\triangle CQB\) by SAS. 4. Corresponding sides of congruent triangles are equal, so \(PD=QB\).

a) SAS, using \(AD=CB\), \(AP=CQ\), and the equal included right angles. b) \(PD=QB\) by corresponding parts of congruent triangles.

- 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 can you say about the two base angles of the isosceles triangle? - Which equal angles are created where the altitude meets the base? - After proving the small triangles congruent, which sides correspond?

1. Since \(AC=BC\), the base angles at \(A\) and \(B\) are equal, so \(\angle CAD=\angle DBC\). 2. Since \(CD\perp AB\), \(\angle ADC\) and \(\angle BDC\) are right angles. 3. The equal sides \(AC\) and \(BC\) are nonincluded relative to those two pairs of angles. Hence \(\triangle ACD\cong\triangle BCD\) by AAS. 4. Corresponding sides give \(AD=BD\).

The equal base angles, right angles at \(D\), and \(AC=BC\) establish \(\triangle ACD\cong\triangle BCD\) by AAS. 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.

- Which two angles does the bisector at \(B\) make equal? - Use the other stated angle equality to match the smaller triangles. - Identify the shared side and then the corresponding outer sides.

1. Because \(BE\) bisects the angle at \(B\), \(\angle ABE=\angle EBC\). 2. The problem also gives \(\angle BAE=\angle BCE\). 3. In triangles \(ABE\) and \(CBE\), the shared side \(BE\) lies opposite the angles at \(A\) and \(C\), so the side is nonincluded. 4. Thus \(\triangle ABE\cong\triangle CBE\) by AAS. Corresponding outer sides give \(AB=CB\).

The angle bisector and the given angles provide two matching angle pairs. With shared nonincluded side \(BE\), AAS proves \(\triangle ABE\cong\triangle CBE\), so \(AB=BC\).

- Match the letters in the order given by the congruence statement. - Which pairs of equal sides are opposite sides of the quadrilateral? - What can two equal pairs of opposite sides tell you about a quadrilateral?

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

\(ABCD\) is a parallelogram, since \(AB=CD\) and \(BC=DA\) by corresponding parts of the congruent triangles.

- 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 \(A\).

- Match the first two letters with the first two letters in the other triangle. - Do the final letters also give a corresponding pair of angles? - Set the matching measurements equal before solving for each variable.

1. From the triangle order, side \(AB\) corresponds to \(PQ\), so \(2x+3=17\) and \(x=7\). 2. Angle \(C\) corresponds to angle \(R\), so \(2y+6=58\) and \(y=26\). 3. Each equation uses equality of corresponding parts of congruent triangles.

\(x=7\) from \(AB=PQ\); \(y=26\) from \(\angle C=\angle R\).

- Locate where each pair of named equal sides meets. - Does the student's angle use those meeting vertices? - Name the angles formed by the two equal sides.

1. In triangle \(ABC\), the sides \(AB\) and \(AC\) meet at \(A\). In triangle \(DEF\), sides \(DE\) and \(DF\) meet at \(D\). 2. The student's given angle pair, \(\angle B=\angle E\), is not between those equal sides. 3. The SAS proof would work with the additional fact \(\angle BAC=\angle EDF\). 4. Without that fact, the stated SAS reasoning is not justified.

No. The given angles are not included between the equal sides. The needed angle pair is \(\angle BAC=\angle EDF\).

- Sketch where the already known angle lies relative to the known side. - For each possible extra fact, list the total number of side and angle pairs. - Which familiar criteria require an included angle or a nonincluded side?

1. Option a) adds a second side pair to the given angle at \(A\), but the known angle is not between \(AB\) and \(BC\). This is SSA, which does not always guarantee congruence. 2. Option b) adds a second corresponding angle pair. The known equal side \(BC=EF\) is not between the two angles at \(A,B\) and \(D,E\). 3. Option b) is therefore sufficient by AAS; option a) is not guaranteed to be sufficient.

a) Not enough in general: SSA. b) Enough: AAS, since \(BC=EF\) is a nonincluded corresponding side.

- Side ticks represent measured lengths; the drawn proportions alone do not prove equality. - Where is the marked angle relative to the two marked sides in each bracket? - After proving congruence, which side connects the vertices corresponding to \(B\) and \(C\)?

1. The first set of ticks gives \(AB=DE\), and the second gives \(AC=DF\). 2. The marked angles \(\angle BAC\) and \(\angle EDF\) are equal and lie between these equal side pairs. 3. Therefore \(\triangle ABC\cong\triangle DEF\) by SAS. 4. The corresponding outer bars \(BC\) and \(EF\) are equal, so \(BC=8\,\text{in.}\).

\(AB=DE\), \(AC=DF\), and \(\angle BAC=\angle EDF\), so \(\triangle ABC\cong\triangle DEF\) by SAS. Thus \(BC=EF=8\,\text{in.}\).

- Start by matching sides of the same length, without assuming the vertex letters match in alphabetical order. - At each vertex, compare the two side lengths that meet there. - After finding the correct vertex order, match the requested angle.

1. The corresponding side pairs are \(AB=EF=5\,\text{cm}\), \(BC=FD=7\,\text{cm}\), and \(CA=DE=6\,\text{cm}\). 2. Vertex \(A\), where the \(5\)- and \(6\)-cm sides meet, matches \(E\). Similarly, \(B\) matches \(F\) and \(C\) matches \(D\). 3. Therefore \(\triangle ABC\cong\triangle EFD\) by SSS. 4. Corresponding angles are equal, so \(\angle ABC=\angle EFD\).

\(AB=EF\), \(BC=FD\), and \(CA=DE\), so \(\triangle ABC\cong\triangle EFD\) by SSS. The angle matching \(\angle ABC\) is \(\angle EFD\).

- What pair of angles does each bisector make equal? - What side belongs to both triangles? - Is that shared side between the two known angle pairs?

1. The bisector at \(A\) gives \(\angle BAD=\angle DAC\). 2. The bisector at \(D\) gives \(\angle BDA=\angle ADC\). 3. Side \(AD\) is shared by triangles \(BAD\) and \(CAD\), and it lies between the two equal angle pairs. 4. Therefore \(\triangle BAD\cong\triangle CAD\) by ASA, so \(AB=AC\) by corresponding parts.

\(\angle BAD=\angle DAC\), \(\angle BDA=\angle ADC\), and \(AD=AD\). By ASA, \(\triangle BAD\cong\triangle CAD\), so \(AB=AC\).

- How many pairs of equal angles can you identify? - Is the equal side between those two angles? - Once the triangles are congruent, which side corresponds to \(AC\)?

1. The two right angles give \(\angle A=\angle D\). 2. Another pair of corresponding angles is given: \(\angle B=\angle E\). 3. The matching equal sides \(BC=EF\) are opposite the right angles, so they are not between the two known angle pairs. 4. Therefore \(\triangle ABC\cong\triangle DEF\) by AAS. Corresponding parts give \(AC=DF\).

\(\angle A=\angle D\), \(\angle B=\angle E\), and the nonincluded sides \(BC=EF\). Thus \(\triangle ABC\cong\triangle DEF\) by AAS and \(AC=DF\).

- Which two smaller triangles share the angle bisector? - Which new equal-angle pair follows from the angle bisector, beyond the angles already given? - Is the shared side between the two angle pairs you used?

1. Since \(D\) lies on \(AB\), \(\angle CAD=\angle DBC\) from the given equal angles at \(A\) and \(B\). 2. Since \(CD\) bisects the angle at \(C\), \(\angle ACD=\angle BCD\). 3. Triangles \(ACD\) and \(BCD\) share side \(CD\). This side is not between the two specified angles, so AAS gives \(\triangle ACD\cong\triangle BCD\). 4. The corresponding outer sides satisfy \(AC=BC\). Therefore triangle \(ABC\) is isosceles.

\(\triangle ACD\cong\triangle BCD\) by AAS, using equal angles at \(A,B\), the bisected angle at \(C\), and shared side \(CD\). Hence \(AC=BC\).

- 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 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 the two equal sides of the inner triangle. - Compare the outer angles at \(D\) and \(E\) with the inner base angles. - Which two outer triangles have two corresponding equal sides and an included angle?

1. Since \(AD=AE\), base angles \(\angle ADE\) and \(\angle DEA\) are equal. 2. As \(B,D,E,C\) lie on one line, \(\angle ADB\) and \(\angle AEC\) are supplementary to those equal base angles. Hence \(\angle ADB=\angle AEC\). 3. In \(\triangle ADB\) and \(\triangle AEC\), \(AD=AE\), \(DB=EC\), and the included angles at \(D\) and \(E\) are equal. 4. SAS gives \(\triangle ADB\cong\triangle AEC\), so \(AB=AC\) by corresponding parts.

Equal base angles of \(ADE\) give equal supplementary angles at \(D\) and \(E\). By SAS, \(\triangle ADB\cong\triangle AEC\), so \(AB=AC\).

- Which angles at the base of the isosceles triangle are equal? - What angles does the altitude create at \(D\)? - Once the two smaller triangles are congruent, what lengths can you work out?

1. Because \(AB=BC\), the base angles \(\angle BAD\) and \(\angle BCD\) are equal. 2. Both angles at \(D\) are right angles because \(BD\) is an altitude, and \(AB=CB\) is given. The equal side is not between the two given angles. 3. Therefore \(\triangle ABD\cong\triangle CBD\) by AAS. 4. Corresponding parts give \(AD=DC=\frac{12}{2}=6\,\text{cm}\). 5. In right triangle \(ABD\), \(BD=\sqrt{10^2-6^2}=\sqrt{64}=8\,\text{cm}\).

\(\triangle ABD\cong\triangle CBD\) by AAS, using the equal base angles, two right angles, and \(AB=BC\). Then \(AD=6\,\text{cm}\) and \(BD=8\,\text{cm}\).

- 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.

- In the two triangles, which angle is opposite their common side? - Distinguish an included side from a side opposite one of the known angles. - After naming corresponding vertices, consider which outer lengths must match.

1. The two smaller triangles have equal angles at \(D\) and equal angles at \(A\) and \(C\), respectively. 2. They share side \(BD\), opposite the corresponding angles at \(A\) and \(C\). 3. Therefore \(\triangle ABD\cong\triangle CBD\) by AAS. 4. Corresponding sides \(AB\) and \(CB\) are equal, so triangle \(ABC\) is isosceles.

\(\triangle ABC\) is isosceles: AAS gives \(\triangle ABD\cong\triangle CBD\), and corresponding sides give \(AB=CB\).

- 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.

- Write one Pythagorean equation for each right triangle. - Which hypotenuse belongs to both triangles? What do the matching tick marks tell you? - What third side equality would let you use SSS?

1. The Pythagorean theorem in right triangle \(ABD\) gives \(BD^2=AD^2-AB^2\). 2. In right triangle \(DCA\), it gives \(CA^2=AD^2-DC^2\). 3. Since \(AB=DC\), the expressions on the right are equal. Hence \(BD^2=CA^2\), so \(BD=CA\) because lengths are nonnegative. 4. We now have \(AB=DC\), \(BD=CA\), and shared hypotenuse \(AD=DA\). Therefore \(\triangle ABD\cong\triangle DCA\) by SSS.

The Pythagorean theorem gives \(BD^2=AD^2-AB^2=AD^2-DC^2=CA^2\), so \(BD=CA\). With \(AB=DC\) and \(AD=DA\), SSS proves congruence.

- Apply the midpoint definition to both crossing segments. - What kind of angles meet at the intersection \(Z\)? - After congruence, match each requested part to its counterpart.

1. Since \(Z\) is the midpoint of both segments, \(AZ=DZ\) and \(BZ=CZ\). 2. Vertical angles \(\angle AZB\) and \(\angle DZC\) are equal. 3. SAS therefore gives \(\triangle ABZ\cong\triangle DCZ\). 4. Corresponding parts yield \(\angle ZDC=\angle ZAB=35^\circ\) 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 equal full side lengths of the outer equilateral triangle. - What lengths remain after removing the equal short pieces from the outer sides? - Which smaller triangles contain the sides of \(DEF\)?

1. Because triangle \(ABC\) is equilateral, \(AB=BC=CA\) and all three angles are \(60^\circ\). 2. Subtracting the equal shorter parts gives \(BD=CE=AF\). 3. In triangles \(ADF\), \(BED\), and \(CFE\), the corresponding side pairs have lengths \(AD=BE=CF\) and \(AF=BD=CE\). 4. Each included angle at an outer vertex is \(60^\circ\), so the three corner triangles are congruent by SAS. 5. Their corresponding third sides are \(DF\), \(DE\), and \(EF\). Thus \(DF=DE=EF\), proving triangle \(DEF\) equilateral.

SAS makes corner triangles \(ADF\), \(BED\), and \(CFE\) congruent. Hence their third sides satisfy \(DF=DE=EF\), so \(DEF\) is equilateral.

- Use a property that every rhombus inherits from being a parallelogram. - Compare the two triangles that share \(\overline{AM}\). - After proving the angles at \(M\) are congruent, use their linear-pair relationship.

1. A rhombus is a parallelogram, so its diagonals bisect each other. Therefore, \(BM=DM\). 2. In triangles \(ABM\) and \(ADM\), \(AB=AD\) because all sides of a rhombus are congruent, \(BM=DM\), and \(AM\) is shared. 3. Thus \(\triangle ABM\cong\triangle ADM\) by SSS, so \(\angle BMA\cong\angle AMD\). 4. These two angles form a linear pair. Congruent supplementary angles each measure \(90^\circ\), so \(AC\perp BD\). 5. Since \(M\) is the midpoint of \(\overline{BD}\), line \(AC\) is the perpendicular bisector of \(\overline{BD}\).

The diagonals of the rhombus bisect each other, so \(BM=DM\). Triangles \(ABM\) and \(ADM\) are congruent by SSS, making the adjacent angles at \(M\) congruent; because they form a linear pair, both are right angles. Therefore, \(\overline{AC}\) is the perpendicular bisector of \(\overline{BD}\).

- Check the reason attached to every claimed angle relationship before checking the final criterion. - Compare Kagiso's three-letter criterion with the valid triangle-congruence criteria in this course. - Look for a side relationship Kagiso 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 marked information gives \(AB = AC\) and \(BD = DC\). 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.

- 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 \(PM\perp AB\).

- Which interior angles are next to the two given exterior angles? - What information do the two pairs of equal sides give you? - After proving congruence, which corresponding parts contain \(t\)?

1. Each given exterior angle forms a straight angle with the included interior angle, so \(\angle BAC=180^\circ-110^\circ=70^\circ\) and \(\angle EDF=70^\circ\). 2. Along with \(AB=DE\) and \(AC=DF\), these equal included interior angles establish \(\triangle ABC\cong\triangle DEF\) by SAS. 3. Corresponding angles \(\angle ABC\) and \(\angle DEF\) are equal. Thus \(2t+10=t+45\), giving \(t=35\). 4. SAS uses the interior angles formed by the given side pairs; the exterior angles use opposite rays and must first be converted to interior angles.

Both included interior angles are \(70^\circ\), so the triangles are congruent by SAS. Then \(2t+10=t+45\) gives \(t=35\). The \(110^\circ\) exterior angles are not the included angles between the named sides.

- Which segments on the two crossing lines at \(M\) are equal? - Identify the pair of vertical angles at \(M\). - Use the outer triangle after the congruence gives a second equal side length.

1. Because \(M\) is the midpoint of \(BC\), \(BM=MC\). Also \(AM=MD\) by the extension condition. 2. Vertical angles \(\angle AMB\) and \(\angle DMC\) are equal. Hence \(\triangle ABM\cong\triangle DCM\) by SAS. 3. Corresponding sides yield \(CD=AB\). 4. The triangle inequality in \(ACD\) gives \(AD<AC+CD\). 5. As \(AD=AM+MD=2AM\) and \(CD=AB\), this becomes \(2AM<AC+AB\). Divide by \(2\) to obtain the claim.

a) \(\triangle ABM\cong\triangle DCM\) by SAS, so \(CD=AB\). b) From \(AD<AC+CD\), use \(AD=2AM\) and \(CD=AB\) to get \(AM<\frac{AC+AB}{2}\).

- Look at the angles next to the two given equal angles on the straight line. - What could you learn from the triangle between \(P\) and \(Q\)? - Which triangles contain the two outer sides you need to compare?

1. Since \(B,P,Q,C\) are on one line, \(\angle APB\) is supplementary to \(\angle APQ\), and \(\angle AQC\) is supplementary to \(\angle AQP\). 2. The given outer angles are equal, so their supplements are equal: \(\angle APQ=\angle AQP\). 3. Equal angles in triangle \(APQ\) have equal opposite sides, giving \(AP=AQ\). 4. In triangles \(ABP\) and \(ACQ\), we now have \(AP=AQ\), \(BP=CQ\), and the equal included angles \(\angle APB=\angle AQC\). 5. SAS gives \(\triangle ABP\cong\triangle ACQ\), so corresponding sides satisfy \(AB=AC\).

The equal given angles have equal supplements at \(P\) and \(Q\), giving \(AP=AQ\). Then SAS proves \(\triangle ABP\cong\triangle ACQ\), so \(AB=AC\).

- 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.

- 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.

- Use the marked side pairs to determine the vertex correspondence before reading Akari'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. Akari'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.