- Count how many pairs of corresponding sides are marked congruent. - No angle information is needed for this criterion.

1. The one-tick sides show \(AB=DE\). 2. The two-tick sides show \(BC=EF\). 3. The three-tick sides show \(AC=DF\). 4. All three pairs of corresponding sides are congruent, so the triangles are congruent by SSS.

SSS

- Identify the two marked angles first. - Ask whether the marked side lies between those two angles.

1. The angle marks show \(\angle G\cong\angle J\) and \(\angle H\cong\angle K\). 2. The side marks show \(GI=JL\). 3. Side \(GI\) is not the side included between the two marked angles \(G\) and \(H\). 4. Therefore, the triangles are congruent by AAS.

AAS

- Count how many corresponding side pairs are marked. - Is the marked angle between those sides or across from one of them?

1. The single ticks give \(AB=DE\), and the double ticks give \(AC=DF\). 2. The equal angles are \(\angle BAC\) and \(\angle EDF\), each between the two marked sides. 3. Two side pairs and their included angles agree, so the criterion is SAS.

SAS. The marked equal angles lie between the two pairs of marked equal sides.

- The location of the equal angle matters, not just its size. - Picture the angle formed by the two measured sides, then compare it with an angle opposite a side.

1. In both triangles, two corresponding sides and the angle included between them agree. 2. That is the SAS congruence criterion. 3. An angle opposite the \(7\,\text{cm}\) side is not included between the given sides. Those measurements are SSA, so SAS does not apply; this alone makes no claim about whether the SSA triangle is unique.

SAS in the first case. No in the second: the angle is not included, so the measurements are SSA, not SAS. This does not imply that two different triangles exist.

- Can a single-arc angle match a double-arc angle? Use the markings to pair the angles. - Check whether the matched angles lie at the ends of a common side in each triangle.

1. The single arcs show \(\angle BAC=\angle CAD\), and the double arcs show \(\angle BCA=\angle ACD\). 2. The common side \(AC\) is between the two marked angles in both triangles. 3. The triangles are congruent by ASA.

ASA, because \(AC\) is the included side between two corresponding equal angles.

- Write down the matching angle vertices first. - Decide whether the measured side connects those two vertices.

1. The equal angle pairs are \(\angle J=\angle M\) and \(\angle K=\angle N\). 2. Side \(JK\) connects vertices \(J\) and \(K\), just as \(MN\) connects \(M\) and \(N\). 3. The known side is included between the known angles, so the direct criterion is ASA.

ASA. The equal sides \(JK\) and \(MN\) lie between the two given equal angles.

- Look at the endpoints of the given side in each triangle. - Are both given angles at those endpoints? - Which congruence criterion fits that placement?

1. Side \(AB\) lies between the two given angles in triangle \(ABC\). 2. The corresponding side \(DE\) lies between the same two angle measures in triangle \(DEF\). 3. Therefore \(\triangle ABC\cong\triangle DEF\) by ASA.

Yes, by ASA. The equal side \(AB=DE\) is included between the two given equal angles.

- Which segment belongs to both triangles? - Make a list of the three sides in each triangle. - What criterion applies if every side has a matching equal side?

1. Both triangles share side \(CD\), so \(CD=DC\) by the reflexive property. 2. The other matching sides are \(AD=BC\) and \(AC=BD\). 3. All three corresponding sides match, so \(\triangle ADC\cong\triangle BCD\) by SSS, with \(A\leftrightarrow B\), \(D\leftrightarrow C\), and \(C\leftrightarrow D\).

a) \(CD=DC\). b) \(\triangle ADC\cong\triangle BCD\) by SSS.

- Subtract the two known side lengths from the perimeter. - Which congruence criterion uses all three side lengths? - Check that the three lengths can form a triangle.

1. Find the third side length: \(20\,\text{cm} - 6\,\text{cm} - 9\,\text{cm} = 5\,\text{cm}\). 2. Both triangles therefore have side lengths \(5\,\text{cm}\), \(6\,\text{cm}\), and \(9\,\text{cm}\). 3. Since all three pairs of corresponding side lengths are equal, the triangles are congruent by SSS.

Yes. Both triangles have side lengths \(5\,\text{cm}\), \(6\,\text{cm}\), and \(9\,\text{cm}\), so they are congruent by SSS.

- Match vertices by their angle measures, not by alphabetical order. - Once two vertices match, what happens to the remaining pair? - Check whether the known side connects the matched angle vertices in each triangle.

1. The \(40^\circ\) angles pair \(A\) with \(E\), and the \(65^\circ\) angles pair \(B\) with \(F\). Therefore \(C\) corresponds to \(D\). 2. The equal sides \(AB\) and \(EF\) are between the given angle pairs. Hence \(\triangle ABC\cong\triangle EFD\) by ASA. 3. Under this correspondence, side \(AC\) matches \(ED\).

\(\triangle ABC\cong\triangle EFD\) by ASA. Side \(AC\) corresponds to \(ED\).

- List the three sides of each triangle. - Identify the side that belongs to both triangles. - Which congruence criterion uses three pairs of sides?

1. The given opposite-side relationships are \(AB = CD\) and \(BC = DA\). 2. The triangles share side \(\overline{AC}\), so \(AC = CA\). 3. Thus, all three pairs of corresponding sides are congruent. Therefore, \(\triangle ABC \cong \triangle CDA\) by SSS.

\(\triangle ABC \cong \triangle CDA\) by SSS because \(AB = CD\), \(BC = DA\), and \(AC = CA\).

- Identify the marked side and angles in each triangle. - Check whether the marked angles are at the endpoints of the marked side. - Rotating or reflecting a triangle does not change congruence. - Which criterion uses two angles and their included side?

1. Triangle 1 has a \(5.0\,\text{cm}\) side with adjacent angles of \(40^\circ\) and \(30^\circ\). 2. Triangle 2 has a \(5.0\,\text{cm}\) side with adjacent angles of \(30^\circ\) and \(40^\circ\). 3. Triangle 3 has sides of \(5.0\,\text{cm}\) and \(3.5\,\text{cm}\) with an included angle of \(40^\circ\). These are not the same given measurements as in the other triangles. 4. Triangle 4 has a \(5.0\,\text{cm}\) side with adjacent angles of \(40^\circ\) and \(30^\circ\). 5. Therefore, triangles 1, 2, and 4 are congruent by ASA. Triangle 3 is not congruent to them.

Triangles 1, 2, and 4 are congruent by ASA.

- Locate each given angle relative to the given sides. - When two sides and an angle are known, is the angle between the sides? - In b), could \(C\) be in more than one place without changing the two lengths or the angle at \(A\)? - For the three-side case, check that the lengths can make a triangle.

1. In a), side \(BC\) lies between the two given angles. This is ASA, a congruence criterion. 2. In b), the angle at \(A\) is opposite side \(BC\), not between the known sides \(AB\) and \(BC\). This is SSA. 3. To see why b) does not fix one triangle, draw a ray from \(A\) making \(30^\circ\) with \(AB\). A circle centered at \(B\) with radius \(4\,\text{cm}\) meets that ray at two points, either of which can be \(C\). The resulting triangles have different lengths \(AC\), so they are not congruent. 4. In c), \(AC\) and \(AB\) meet at the given angle at \(A\). This is SAS, a congruence criterion. 5. In d), all three side lengths are given. The strict triangle inequalities hold, so this is SSS, a congruence criterion.

a) ASA — sufficient. b) SSA — insufficient: a circle of radius \(4\,\text{cm}\) centered at \(B\) meets the \(30^\circ\) ray from \(A\) at two possible points \(C\). c) SAS — sufficient. d) SSS — sufficient; the side lengths can form a triangle.

- Identify the two equal distances and the angle at the trail junction. - Is the marked angle between the two known sides? - How do the two cabin positions compare along Trail 2?

1. The shared path is \(AB=8\,\text{mi}\). Both cabins are \(5\,\text{mi}\) from \(B\), so \(BC_1=BC_2=5\,\text{mi}\). The angle at \(A\) is \(30^\circ\) in each triangle. 2. The angle at \(A\) is opposite the known \(BC\) side, rather than between \(AB\) and \(BC\), so this is SSA. 3. The two cabins are at different distances from \(A\) along the same trail, so \(AC_1\ne AC_2\). Corresponding sides therefore differ, and the triangles are not congruent.

a) \(AB\) is shared, \(BC_1=BC_2=5\,\text{mi}\), and both angles at \(A\) are \(30^\circ\). b) They are not congruent because \(AC_1\ne AC_2\). SSA does not guarantee congruence.

- Compare all the given measurements. - Which congruence theorem applies when all three corresponding sides are equal? - For the angle-side-angle data, check that the angle measures match exactly.

1. Triangles (1) and (3) each have side lengths \(3\,\text{cm}\), \(5\,\text{cm}\), and \(6\,\text{cm}\). 2. Therefore, triangles (1) and (3) are congruent by SSS. 3. Triangles (2) and (4) each have a side of \(4.5\,\text{cm}\) and one angle of \(35^{\circ}\), but their other given angles are different: \(105^{\circ}\) and \(100^{\circ}\). They are not congruent. 4. Neither triangle (2) nor triangle (4) can be congruent to triangle (1) or (3), because the given side length \(4.5\,\text{cm}\) does not match any side length in those triangles.

Only triangles (1) and (3) are congruent, by SSS. No other pair is congruent.

- Identify the side shared by both triangles. - List the other marked equal side and angle. - Is the equal angle included between the two equal sides?

1. The markings give \(KN = KL\). 2. The triangles share side \(\overline{KM}\), so \(KM = KM\). 3. The marked angles satisfy \(\angle NKM = \angle LKM\), and each is included between the corresponding equal sides. 4. Therefore, \(\triangle KNM \cong \triangle KLM\) by SAS.

The triangles are congruent by SAS.

- Use the equal legs of the isosceles triangle. - State what an angle bisector does. - Identify the shared side.

1. The given equal sides are \(PQ = PR\). 2. Because \(\overline{PS}\) bisects \(\angle QPR\), \(\angle QPS = \angle RPS\). 3. The triangles share side \(\overline{PS}\), so \(PS = PS\). 4. Therefore, \(\triangle PQS \cong \triangle PRS\) by SAS.

\(\triangle PQS \cong \triangle PRS\) by SAS.

- Recall the side properties of a rectangle. - Compare the included angles at \(B\) and \(D\). - Look for a second valid criterion using the diagonal.

1. Opposite sides of a rectangle are congruent, so \(AB = CD\) and \(BC = DA\). 2. Angles \(\angle ABC\) and \(\angle CDA\) are both right angles. 3. Therefore, \(\triangle ABC \cong \triangle CDA\) by SAS. 4. Alternatively, the triangles also share \(\overline{AC}\), so SSS can be used.

The triangles are congruent by SAS. SSS is also valid.

- State what midpoint means. - Identify the shared side. - Check whether the equal angles are included between the equal sides.

1. Since \(M\) is the midpoint of \(\overline{AB}\), \(AM = BM\). 2. The triangles share side \(\overline{MC}\), so \(MC = MC\). 3. The included angles satisfy \(\angle AMC = \angle BMC = 90^\circ\). 4. Therefore, \(\triangle AMC \cong \triangle BMC\) by SAS.

The triangles are congruent by SAS.

- The three angles of triangle \(DEF\) sum to \(180^\circ\). - Compare the angles at the endpoints of \(BC\) and \(EF\). - Does the known side lie between the matching angle pairs?

1. In \(DEF\), \(\angle F=180^\circ-65^\circ-47^\circ=68^\circ\). 2. Thus \(\angle B=\angle E=47^\circ\) and \(\angle C=\angle F=68^\circ\). 3. The equal side \(BC=EF=8.3\,\text{cm}\) joins the two matching angle vertices in each triangle. Therefore \(\triangle ABC\cong\triangle DEF\) by ASA.

a) \(\angle F=68^\circ\) b) \(\triangle ABC\cong\triangle DEF\) by ASA, using \(\angle B=\angle E\), \(\angle C=\angle F\), and \(BC=EF\).

- What distances equal the radius of each circle? - Which side belongs to both triangles? - What does a reflection preserve, and what happens to points on its mirror line?

1. Since \(R\) and \(S\) lie on each circle, \(PR=PS\) and \(QR=QS\). The triangles share \(PQ\). 2. Hence \(\triangle PRQ\cong\triangle PSQ\) by SSS. 3. The two triangles have the same three side lengths, share base \(PQ\), and lie on opposite sides of that base. 4. Reflection across \(PQ\) fixes \(P\) and \(Q\) and takes the third vertex of one triangle to the third vertex of the other, so it maps \(R\) to \(S\).

a) SSS, since \(PR=PS\), \(QR=QS\), and \(PQ=QP\). b) The congruent triangles are on opposite sides of their shared base \(PQ\); reflection across that line exchanges \(R\) and \(S\).

- Identify the vertex where the two marked sides meet in each triangle. - SAS uses the angle between the two known side pairs. - Write each angle name with its vertex in the middle.

1. The two given side pairs meet at \(A\) in the first triangle and at \(D\) in the second triangle. 2. SAS requires the included angles between those pairs of sides. 3. Therefore, the needed additional fact is \(\angle BAC\cong\angle EDF\). 4. With \(AB = DE\), \(AC = DF\), and \(\angle BAC\cong\angle EDF\), the triangles are congruent by SAS.

\(\angle BAC\cong\angle EDF\)

- Match the angle marks before matching the side marks. - Decide whether the marked side is included between the two marked angles. - After proving congruence, identify which side corresponds to \(BC\).

1. The angle marks show \(\angle A\cong\angle D\) and \(\angle B\cong\angle E\). 2. The tick marks show \(AC=DF\). This side is not included between the two marked angles. 3. Therefore, \(\triangle ABC\cong\triangle DEF\) by AAS. 4. Corresponding sides are congruent, so \(BC=EF=7\,\text{cm}\).

The triangles are congruent by AAS, and \(BC=7\,\text{cm}\).

- In each panel, locate the two marked angles first. - Then ask whether the marked side connects the vertices of those two angles. - “Included” means the side lies between the two given angles.

1. In panel a), the marked side lies between the two marked angles in each triangle, so the criterion is ASA. 2. In panel b), the marked side is not between the two marked angles, so the criterion is AAS.

a) ASA b) AAS

- Find the vertex shared by the two sides \(AB\) and \(BC\). - The matching sides in the other triangle must share its corresponding vertex. - Read the three letters in the same corresponding order.

1. Since \(AB\) matches \(DE\) and \(BC\) matches \(EF\), vertex \(B\) must match the common vertex \(E\). 2. The remaining correspondences are \(A\leftrightarrow D\) and \(C\leftrightarrow F\). 3. The correct statement is \(\triangle ABC\cong\triangle DEF\) by SSS. 4. The student's statement pairs \(AB\) with \(DF\), not the given corresponding side \(DE\).

\(\triangle ABC\cong\triangle DEF\). The incorrect statement matches \(AB\) with \(DF\) instead of \(DE\).

- Match the vertices in the order of the required congruence statement. - Decide which angle belongs with vertex \(B\). - Ask whether the supplied equal side lies between the two known angles.

1. The intended correspondence is \(A\leftrightarrow D\), \(B\leftrightarrow E\), \(C\leftrightarrow F\). 2. Adding \(\angle B=\angle E\) supplies two corresponding angle pairs. 3. The equal side \(BC=EF\) is not between the known angles at \(A\) and \(B\), so AAS applies. 4. \(\angle B=\angle F\) would not follow the requested vertex order.

\(\angle B=\angle E\). Side \(BC\) is not the included side between \(\angle A\) and \(\angle B\), so AAS proves the congruence.

- Congruent triangles must agree in both shape and size. - Check the side lengths before drawing a conclusion from the angles. - What additional equal measurements would distinguish congruence from merely having the same angles?

1. The two triangles have the same three angle measures. 2. They are not congruent because one triangle has sides \(3\,\text{cm}\) long and the other has sides \(6\,\text{cm}\) long. 3. Three equal angles alone establish the same shape, but not the same size. 4. Equal lengths for all three pairs of corresponding sides would instead permit SSS.

No. Their side lengths differ, so the triangles are different sizes. Equal corresponding sides would allow SSS.

- Match the pairs of marked angles before interpreting the marked side. - Where does the ticked side lie relative to the two marked angles? - Once the triangles are congruent, what can you say about their other corresponding sides?

1. The single arcs show \(\angle A=\angle D\), and the double arcs show \(\angle B=\angle E\). 2. The matching ticks show \(AB=DE\), the sides included between the corresponding marked angles. 3. Therefore \(\triangle ABC\cong\triangle DEF\) by ASA. 4. Corresponding sides of congruent triangles are equal, so \(BC=EF\).

Yes. \(\triangle ABC\cong\triangle DEF\) by ASA, so \(BC=EF\).

- Work out the third angle before comparing the side positions. - Which side is between the two given angles? Which side is opposite one of them? - For congruence, can the same stated length correspond to sides opposite different angles?

1. The third angle of each triangle is \(180^\circ-50^\circ-70^\circ=60^\circ\). 2. In \(ABC\) and \(DEF\), the given side joins the vertices with the \(50^\circ\) and \(70^\circ\) angles. The data use ASA, so \(\triangle ABC\cong\triangle DEF\). 3. In \(GHI\), side \(HI\) is opposite the \(50^\circ\) angle, rather than between the given angles. Its information uses AAS. 4. Congruent triangles would have to match vertices with equal angles because all three angles are different. Thus \(HI\) corresponds to \(BC\), not \(AB\). In \(ABC\), sides \(BC\) and \(AB\) cannot be equal because their opposite angles are \(50^\circ\) and \(60^\circ\). 5. Therefore \(GHI\) is not congruent to the other two triangles.

\(\triangle ABC\cong\triangle DEF\) by ASA. The givens for \(GHI\) use AAS, but its \(6\,\text{cm}\) side lies opposite a different angle, so it is not congruent to the other two.

- First express the two side lengths in the same unit. - What is the sum of the three interior angles of a triangle? - Which equal angles are at the endpoints of each given side?

1. Convert \(HI=4.2\,\text{cm}=42\,\text{mm}\), so \(BC=HI\). 2. In \(ABC\), \(\angle A=180^\circ-100^\circ-35^\circ=45^\circ\). 3. In \(GHI\), \(\angle I=180^\circ-100^\circ-45^\circ=35^\circ\). 4. The equal sides \(BC\) and \(HI\) are included between the equal \(100^\circ\) and \(35^\circ\) angles. Hence \(\triangle ABC\cong\triangle GHI\) by ASA.

Yes. \(HI=42\,\text{mm}\), \(\angle A=45^\circ\), and \(\angle I=35^\circ\). The triangles are congruent by ASA.

- Test the triangle inequality separately on either side of the diagonal. - What happens if two of a triangle's side lengths add to the third? - When all three side lengths of a triangle are fixed, what can vary besides its position or reflection?

1. If \(AC=10\,\text{cm}\), triangle \(ABC\) has side lengths \(7,5,10\), satisfying the strict triangle inequalities. 2. Triangle \(ADC\) would have sides \(6,4,10\). Since \(6+4=10\), its vertices would be collinear, so that diagonal is impossible for a quadrilateral. 3. If \(AC=8\,\text{cm}\), triangle \(ABC\) has sides \(7,5,8\), and triangle \(ADC\) has sides \(6,4,8\). Both satisfy the strict triangle inequalities. 4. Each triangle is fixed up to congruence by SSS. Placing their third vertices on opposite sides of \(AC\) gives the convex quadrilateral, determined up to reflection.

a) No. \(6+4=10\), so triangle \(ADC\) would be degenerate. b) Yes. Triangles \(ABC\) and \(ADC\) both exist and are each determined by SSS; opposite-side placement fixes the convex quadrilateral up to congruence.

- Where do the two known sides meet in each triangle? - What else must match before two sides establish SAS? - Once the triangles match, which vertices and sides correspond?

1. The equal sides meet at \(A\) in the first triangle and at \(D\) in the second. Their included angles are \(90^\circ\) and \(60^\circ\), so SAS cannot establish congruence. In fact the triangles are not congruent with these side correspondences: their included angles differ. 2. Change \(\angle EDF\) to \(90^\circ\). Then \(AB=DE\), \(AC=DF\), and the included angles at \(A\) and \(D\) agree. Therefore \(\triangle ABC\cong\triangle DEF\) by SAS. 3. The matching vertices are \(A\leftrightarrow D\), \(B\leftrightarrow E\), and \(C\leftrightarrow F\). Therefore \(BC\) matches \(EF\), and \(BC=EF\) by corresponding parts of congruent triangles.

a) No. The angles between the matching side pairs differ: \(90^\circ\) and \(60^\circ\). b) Change \(\angle EDF\) to \(90^\circ\). The triangles are then congruent by SAS. c) \(EF=BC\), since \(B\leftrightarrow E\) and \(C\leftrightarrow F\).

- How do you find the perimeter of a parallelogram from two adjacent side lengths? - In the second case, which two sides meet at the given angle? - For the diagonal case, first check that three side lengths can form a triangle.

1. The perimeter is already \(2(5+3)=16\,\text{cm}\). Thus part a) adds no information about the angle, and many parallelograms remain possible. 2. In part b), \(AB\), \(AD\), and their included angle fix \(\triangle ABD\) by SAS. The remaining vertex of the parallelogram is then fixed, up to congruence. 3. In part c), \(AB=5\,\text{cm}\), \(AD=3\,\text{cm}\), and \(BD=7\,\text{cm}\) satisfy the strict triangle inequalities because \(5+3>7\). 4. SSS fixes \(\triangle ABD\). The parallelogram's fourth vertex is then determined, up to congruence.

a) No; the perimeter follows from the two given sides. b) Yes; SAS fixes \(\triangle ABD\). c) Yes; SSS fixes \(\triangle ABD\), since its side lengths can form a triangle.

- Which known sides and angle meet at \(A\)? - Once \(A\), \(B\), and \(D\) are fixed, what line must contain \(C\)? - Use the length of the shorter base and the vertex order to select the location.

1. In triangle \(ABD\), the sides \(AB=7\,\text{cm}\) and \(AD=4\,\text{cm}\) include \(\angle BAD=70^\circ\). These data fix triangle \(ABD\) by SAS. 2. The line through \(D\) parallel to \(AB\) contains \(C\). The ordered convex trapezoid places \(C\) in the direction from \(A\) toward \(B\) at distance \(DC=3\,\text{cm}\). 3. That fixes \(C\) uniquely, so the trapezoid is fixed up to congruence as well.

a) SAS, using \(AB=7\,\text{cm}\), \(AD=4\,\text{cm}\), and included \(\angle BAD=70^\circ\). b) The parallel base from \(D\) of length \(3\,\text{cm}\) fixes \(C\) in the specified vertex order.

- What is the angle sum of two consecutive angles in a parallelogram? - Which angles are opposite, and which sides are opposite? - Locate the angle included between each pair of matching sides.

1. Consecutive angles of a parallelogram are supplementary, so \(\angle ABC=180^\circ-60^\circ=120^\circ\). 2. Opposite angles are equal, giving \(\angle CDA=120^\circ\). 3. Opposite sides are equal: \(AB=CD=5\,\text{cm}\) and \(BC=DA=3\,\text{cm}\). 4. The included angles between those pairs of corresponding sides are \(\angle ABC\) and \(\angle CDA\), each \(120^\circ\). Hence the triangles are congruent by SAS.

a) \(\angle ABC=\angle CDA=120^\circ\) b) \(AB=CD\), \(BC=DA\), and the included angles at \(B\) and \(D\) are equal, so SAS proves congruence.

- Find the third angle in each triangle. - Does the longest side of a right triangle correspond to a leg? - Which side lengths must match when you apply ASA or AAS?

1. Both triangles are right triangles because their third angle is \(180^\circ-45^\circ-45^\circ=90^\circ\). 2. In the first triangle, the two legs are each \(6\,\text{cm}\), so the hypotenuse is \(6\sqrt{2}\,\text{cm}\). 3. In the second, the hypotenuse is \(6\,\text{cm}\), so its two legs are each \(3\sqrt{2}\,\text{cm}\). 4. The corresponding sides therefore have different lengths, so the triangles are not congruent. ASA or AAS requires the given equal side to have the same corresponding position relative to the given angles.

No. A \(6\,\text{cm}\) leg is not the corresponding side to a \(6\,\text{cm}\) hypotenuse. Matching angles and a length do not ensure congruence unless the length belongs to corresponding sides.

- Use the triangle angle-sum theorem. - Must the \(6\,\text{cm}\) side lie between the two stated angles? - List the angle that could be opposite the given side in each case.

1. The third angle is \(180^\circ - 45^\circ - 75^\circ = 60^\circ\). 2. The \(6\,\text{cm}\) side can be opposite the \(45^\circ\), \(60^\circ\), or \(75^\circ\) angle. 3. In each case, the two angles at the endpoints of the \(6\,\text{cm}\) side are fixed, so ASA determines one triangle. 4. The three triangles are not congruent because the same \(6\,\text{cm}\) side is opposite a different angle in each case. Therefore, Lucas is correct.

Lucas is correct. The third angle is \(60^\circ\), and the \(6\,\text{cm}\) side can be opposite the \(45^\circ\), \(60^\circ\), or \(75^\circ\) angle. These three placements produce three noncongruent triangles.

- Check whether the three sides in a) can meet to close a triangle. - Could a dilation change a triangle's size without changing its angles? - In c), is the known angle between the two known sides?

1. In a), \(AC+BC=3+5=8\,\text{cm}<AB=9\,\text{cm}\). The triangle inequality fails, so no triangle exists. 2. In b), the angles fix no side length. A dilation with any positive scale factor gives another triangle with the same angles but different side lengths, so there are infinitely many noncongruent possibilities. 3. In c), \(AB\) and \(BC\) meet at \(B\); the \(45^\circ\) angle between them is included. SAS determines one triangle up to congruence.

a) No triangle exists: \(3+5<9\). b) Not unique: the angles stay the same under a dilation that changes lengths. c) Yes, SAS gives a unique triangle up to congruence.

- After fixing one side and one angle, where is the third vertex allowed to lie? - How can a compass circle represent the fixed length of the side opposite the given angle? - Think about how many times a circle can intersect a ray.

1. Draw the fixed side \(\overline{AB}\) and the ray from \(A\) determined by the given angle. 2. The third vertex \(C\) must lie somewhere on that ray. 3. The fixed length of the side opposite \(\angle A\) means \(C\) must also lie on a circle centered at \(B\) with that side length as its radius. 4. For some valid SSA measurements, the circle intersects the angle ray at two distinct points. 5. Those two intersection points produce two different triangles with the same two side lengths and the same nonincluded angle. Therefore, SSA does not determine a unique triangle and is not a general congruence criterion.

The fixed angle determines a ray from \(A\), while the fixed opposite-side length determines a circle centered at \(B\). If that circle meets the ray at two points, either intersection can be the third vertex. The two resulting triangles satisfy the same SSA data but are not congruent, so SSA is not a general congruence criterion.

- Which segment is shared by the two triangles formed by the diagonal? - Find three lengths for each triangle, counting the shared one only once. - Why does the convex condition matter when putting the triangles together?

1. Choose the four side lengths \(AB\), \(BC\), \(CD\), \(DA\), together with diagonal length \(AC\). 2. Triangle \(ABC\) is fixed by \(AB\), \(BC\), and \(AC\) using SSS. 3. Triangle \(ADC\) is fixed by \(AD\), \(DC\), and \(AC\) using SSS. 4. In a convex quadrilateral, \(B\) and \(D\) lie on opposite sides of the diagonal. Joining the two determined triangles on opposite sides fixes the quadrilateral up to congruence.

Choose \(AB\), \(BC\), \(CD\), \(DA\), and \(AC\). They determine each triangle formed by diagonal \(AC\) by SSS, fixing the convex quadrilateral up to congruence.

- List the three side lengths of each triangle. - Include the shared diagonal. - Use the order of the congruence statement to match corresponding vertices.

1. The triangles have \(AB = CD = 5\,\text{cm}\) and \(BC = DA = 3\,\text{cm}\). 2. They share diagonal \(\overline{AC}\), so \(AC = CA = 6\,\text{cm}\). 3. Therefore, \(\triangle ABC \cong \triangle CDA\) by SSS. 4. The vertex correspondence is \(A \leftrightarrow C\), \(B \leftrightarrow D\), and \(C \leftrightarrow A\). 5. Thus, \(\angle BAC\) corresponds to \(\angle DCA\), and \(\angle BCA\) corresponds to \(\angle DAC\).

a) The triangles are congruent by SSS. b) \(\angle BAC\) corresponds to \(\angle DCA\), and \(\angle BCA\) corresponds to \(\angle DAC\).

- Identify which angle lies between the two known sides in each plan. - Compare the information in each plan with the standard triangle-congruence criteria. - Ask whether the third vertex could satisfy the second plan in more than one position.

1. Plan A copies two sides and the included angle between them: \(AB\), \(AC\), and \(\angle BAC\). 2. These data determine a congruent triangle by SAS, so Plan A guarantees a reproduction of \(\triangle ABC\), up to reflection. 3. Plan B fixes \(AB\), \(AC\), and \(\angle ABC\). The copied angle is not the included angle between the two known sides. 4. This SSA information can produce two different positions for the third vertex, so it does not guarantee a unique congruent triangle.

a) Plan A. b) Plan A uses SAS. Plan B uses SSA, which can be ambiguous and therefore does not guarantee a congruent reproduction.

- Find the third angle using the triangle angle sum. - Which of the originally known angles is opposite \(AB\)? - After finding the missing angle, which two known angles meet at \(AB\)?

1. The angle sum gives \(\angle B=180^\circ-48^\circ-67^\circ=65^\circ\). 2. In the original givens, \(AB\) lies opposite the given angle at \(C\), not between the two given angles at \(A\) and \(C\). Thus AAS determines the triangle up to congruence. 3. Once \(\angle B\) is known, side \(AB\) lies between the known angles at \(A\) and \(B\). Therefore ASA is another valid criterion.

a) \(\angle B=65^\circ\) b) The original givens use AAS. After finding \(\angle B\), the equal data can be viewed as ASA with side \(AB\) included between \(\angle A\) and \(\angle B\).

- Keep the stated vertex correspondence fixed in every part. - For cases with an angle and two sides, check whether the angle is included. - For cases with two angles and a side, check whether the side is included.

1. In a), all three corresponding sides are given, so SSS guarantees congruence. 2. In b), the given angle is included between the two given sides, so SAS guarantees congruence. 3. In c), the given side is included between the two given angles, so ASA guarantees congruence. 4. In d), the given side is not included between the two given angles, so AAS guarantees congruence. 5. In e), the angle is not included between the two given sides. This is SSA, which does not guarantee congruence.

a) Yes — SSS b) Yes — SAS c) Yes — ASA d) Yes — AAS e) No — SSA is not a congruence criterion

- First use one equal side to place two corresponding vertices on top of each other. - After aligning \(A\) with \(D\) and \(B\) with \(E\), what distances must the image of \(C\) have from those two fixed points? - How are the two possible points with those two distances related to line \(DE\)?

1. Translate \(A\) to \(D\). 2. Rotate about \(D\) until \(B\) maps to \(E\). This is possible because \(AB=DE\), and the translation and rotation are rigid motions. 3. After those motions, the image of \(C\) is a point whose distances from \(D\) and \(E\) equal \(DF\) and \(EF\), respectively. 4. There are at most two such locations, on opposite sides of line \(DE\). If the image of \(C\) is already \(F\), the triangles coincide. Otherwise, reflect across line \(DE\); points \(D\) and \(E\) stay fixed and the other possible location maps to \(F\). 5. Thus, a sequence of rigid motions maps \(\triangle ABC\) onto \(\triangle DEF\), which explains why three corresponding side lengths determine congruence.

Translate \(A\) to \(D\), rotate so \(B\) maps to \(E\), and, if needed, reflect across \(DE\) so the image of \(C\) maps to \(F\). The marked side equalities force the third vertex to one of those two mirror-image positions, so a rigid-motion sequence superposes the triangles.

- Think of a four-sided figure that can flex while its side lengths remain fixed. - How do a square and a non-square rhombus differ? - Divide the quadrilateral with a diagonal and count the information needed for each triangle.

1. A square with four side lengths of \(5\,\text{cm}\) has four right angles. A non-square rhombus can also have four side lengths of \(5\,\text{cm}\), but it has acute and obtuse angles. 2. The square and rhombus have the same four side lengths but are not congruent. Therefore, four side lengths do not determine a quadrilateral. 3. In general, five suitable independent measurements are needed. 4. Draw a diagonal to divide the quadrilateral into two triangles. Three suitable measurements determine the first triangle and therefore determine the diagonal. The second triangle shares that diagonal and needs two additional suitable measurements. Thus, the total is \(3 + 2 = 5\).

a) SSSS is not a quadrilateral congruence criterion because a \(5\,\text{cm}\) square and a non-square \(5\,\text{cm}\) rhombus have the same side lengths but different angles. b) An arbitrary quadrilateral generally requires five suitable independent measurements, found by determining two triangles that share a diagonal.

- In a rectangle, what type of triangle is formed by two adjacent sides and a diagonal? - Which side must be longest in a right triangle? - How do the diagonals of a parallelogram intersect?

1. If \(ABCD\) were a rectangle, \(\angle ABC\) would be a right angle, so \(AC\) would be the hypotenuse of right triangle \(ABC\). 2. The hypotenuse must be the longest side of a right triangle. Here, \(AB = 6\,\text{cm}\) is longer than \(AC = 5\,\text{cm}\), so \(AC\) cannot be the hypotenuse. Therefore, the quadrilateral cannot be a rectangle. 3. If the figure is a parallelogram, triangle \(ABC\) is determined by side lengths \(6\,\text{cm}\), \(4\,\text{cm}\), and \(5\,\text{cm}\) using SSS. 4. The diagonals of a parallelogram bisect each other, so \(D\) is the image of \(B\) under a \(180^\circ\) rotation about the midpoint of \(AC\). Thus, the parallelogram is uniquely determined up to congruence.

a) It cannot be a rectangle because \(AC\) would have to be the hypotenuse of \(\triangle ABC\), but \(AB = 6\,\text{cm}\) is longer than \(AC = 5\,\text{cm}\). b) Yes. Triangle \(ABC\) is determined by SSS, and the parallelogram condition fixes \(D\) as the half-turn image of \(B\) about the midpoint of \(AC\).

- Begin by making one pair of corresponding vertices coincide. - Use one equal side to align a second vertex with a rotation. - If the third vertex ends up on the opposite side of the aligned side, what rigid motion can fix that side pointwise? - Then use the equal included angle and the remaining equal side to locate the third vertex.

1. Translate \(\triangle DEF\) so that \(D\) lands on \(A\). 2. Rotate the translated triangle about \(A\) until the image of \(E\) lies on ray \(AB\). Because \(DE = AB\), the image of \(E\) lands on \(B\). 3. The image of ray \(DF\) now makes the same angle with line \(AB\) as ray \(AC\). If the image of \(F\) is on the opposite side of \(AB\) from \(C\), reflect across line \(AB\); this reflection fixes \(A\) and \(B\). 4. After that optional reflection, the image of ray \(DF\) coincides with ray \(AC\). 5. Because \(DF = AC\), the image of \(F\) lands on \(C\). 6. Thus, a sequence of rigid motions maps \(\triangle DEF\) exactly onto \(\triangle ABC\). By the rigid-motion definition of congruence, the triangles are congruent.

Translate \(D\) to \(A\), rotate so that \(E\) lands on \(B\), and, if necessary, reflect across \(AB\) so the third vertex lies on the same side of \(AB\) as \(C\). The equal included angle aligns the third ray, and \(DF = AC\) makes \(F\) land on \(C\). Therefore, a sequence of rigid motions maps one triangle exactly onto the other, so the triangles are congruent.

- Start by superposing the one pair of equal sides. - Once the endpoints of that side coincide, what does each given angle tell you about the direction to the third vertex? - Two fixed rays from the endpoints can meet in how many points on a chosen side of the base?

1. Translate \(A\) to \(D\), then rotate about \(D\) until \(B\) maps to \(E\). This is possible because \(AB=DE\). 2. After \(A\) and \(B\) have been aligned with \(D\) and \(E\), the marked angle at \(A\) fixes the ray on which the image of \(C\) must lie from \(D\), and the marked angle at \(B\) fixes the ray on which it must lie from \(E\). 3. If those rays lie on the opposite side of \(DE\) from the target triangle, reflect across \(DE\). This keeps \(D\) and \(E\) fixed while putting the rays on the target side. 4. Two nonparallel rays from \(D\) and \(E\) have only one intersection. That intersection is \(F\), so the image of \(C\) must be \(F\). 5. Therefore, a sequence of rigid motions superposes the triangles, explaining why the included side and its two endpoint angles determine congruence.

Align the marked side by a translation and rotation, reflect across that side if necessary, and use the two marked endpoint angles to fix two rays whose unique intersection is the third vertex. The resulting rigid-motion sequence maps \(\triangle ABC\) onto \(\triangle DEF\).

- Before moving either triangle, determine what the two marked angle pairs imply about the unmarked third angles. - Which marked side can you superpose by a translation and rotation? - After that side is aligned, use an angle at each endpoint to locate the remaining vertex.

1. The two marked angle equalities imply the third angles are also congruent, because the angle measures in each triangle sum to \(180^\circ\). Thus, \(\angle C\cong\angle F\). 2. Translate \(A\) to \(D\), then rotate about \(D\) until \(C\) maps to \(F\). This is possible because the marked sides give \(AC=DF\). 3. After \(A\) and \(C\) are aligned with \(D\) and \(F\), the marked angle at \(A\) fixes the ray from \(D\) toward the third vertex, and the derived equal angle at \(C\) fixes the ray from \(F\). 4. Reflect across \(DF\) if necessary to place those rays on the same side as \(E\). Their unique intersection is \(E\), so the image of \(B\) is \(E\). 5. Therefore, a sequence of rigid motions maps \(\triangle ABC\) onto \(\triangle DEF\), explaining why the AAS data determine congruence.

First use the triangle angle sum to obtain the third angle pair. Then align the marked nonincluded side by a translation and rotation, reflect across it if needed, and use the two endpoint-angle rays to force the remaining vertex. This rigid-motion sequence superposes the triangles.