Free math worksheets

- 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

- Identify where the given side lies relative to the two given angles. - Match the corresponding parts of the two triangles. - Which criterion uses two angles and the included side?

1. In \(\triangle ABC\), side \(c\) lies between angles \(\alpha\) and \(\beta\). 2. In \(\triangle DEF\), side \(f\) lies between angles \(\delta\) and \(\epsilon\). 3. The corresponding information matches: \(c = f = 5.4\,\text{cm}\), \(\alpha = \delta = 42^\circ\), and \(\beta = \epsilon = 76^\circ\). 4. Therefore, the triangles are congruent by ASA.

Yes. The triangles are congruent by ASA.

- Compare the vertex names of the two triangles. - Match the equal legs and equal diagonals. - Which criterion uses three pairs of congruent sides?

1. Both triangles contain side \(\overline{CD}\). 2. In \(\triangle ADC\), the three sides are \(AD = 5\,\text{cm}\), \(AC = 7\,\text{cm}\), and \(CD\). 3. In \(\triangle BCD\), the three sides are \(BC = 5\,\text{cm}\), \(BD = 7\,\text{cm}\), and \(CD\). 4. Since \(AD = BC\), \(AC = BD\), and \(CD\) is shared, the triangles are congruent by SSS.

a) The shared side is \(\overline{CD}\). b) The triangles are congruent by SSS because \(AD = BC\), \(AC = BD\), and \(CD = CD\).

- 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.

- Begin with the given side and treat it as the included side between the two given angles. - Describe where each angle ray must start and on which side of the base the rays should meet. - Which congruence criterion uses two angles and their included side?

1. Draw \(\overline{AB}\) with \(AB = 7.5\,\text{cm}\). 2. At \(A\), draw a ray that forms a \(40^\circ\) angle with \(\overline{AB}\). 3. At \(B\), draw a ray that forms a \(65^\circ\) angle with \(\overline{BA}\) on the same side of \(\overline{AB}\). 4. Label the intersection of the two rays \(C\). This determines \(\triangle ABC\). 5. The side \(\overline{AB}\) and the two angles at its endpoints are fixed, so every such triangle is congruent by ASA.

a) Draw \(AB = 7.5\,\text{cm}\), construct a \(40^\circ\) ray at \(A\) and a \(65^\circ\) ray at \(B\) on the same side of \(AB\), and label their intersection \(C\). b) Every triangle with these data is congruent by ASA.

- 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.

- 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.

- 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.

- 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 third angle in each triangle. - For each description, ask whether the given side lies between the two given angles. - Because all three angle measures differ, the angle correspondence is forced. - If two sides of one triangle were equal, what would have to be true about their opposite angles?

1. In each triangle the third angle is \(180^\circ-50^\circ-70^\circ=60^\circ\). 2. For Lucas and Sophie, the given \(6\,\text{cm}\) side lies between the \(50^\circ\) and \(70^\circ\) angles, so their data use ASA. Thus \(\triangle ABC\cong\triangle DEF\). 3. Jonah's side \(g=HI\) is opposite the \(50^\circ\) angle, so his two angles and side form AAS, not ASA. 4. Because all three angle measures are different, any congruence with Lucas's triangle would have to match the \(50^\circ\), \(70^\circ\), and \(60^\circ\) vertices respectively. Jonah's \(6\,\text{cm}\) side would then correspond to side \(a\) of Lucas's triangle, while Lucas's given \(6\,\text{cm}\) side is \(c\). 5. In Lucas's triangle, \(a\ne c\); otherwise equal sides \(a\) and \(c\) would force their opposite angles \(A\) and \(C\) to be equal, but \(50^\circ\ne60^\circ\). Therefore Jonah's triangle is not congruent to Lucas's or Sophie's triangle.

Lucas's and Sophie's triangles are congruent by ASA. Jonah's data use AAS, but his \(6\,\text{cm}\) side occupies a different forced correspondence position, so his triangle is not congruent to theirs.

- Convert the side lengths to the same unit. - Use the triangle angle sum to find each missing angle. - Identify the two angles adjacent to the given side in each triangle.

1. Convert the side length: \(4.2\,\text{cm} = 42\,\text{mm}\). 2. In \(\triangle ABC\), \(\alpha = 180^\circ - 100^\circ - 35^\circ = 45^\circ\). 3. In \(\triangle GHI\), \(\angle I = 180^\circ - 100^\circ - 45^\circ = 35^\circ\). 4. Side \(a = 42\,\text{mm}\) lies between the \(100^\circ\) and \(35^\circ\) angles. Side \(g = 42\,\text{mm}\) also lies between the \(100^\circ\) and \(35^\circ\) angles. 5. Therefore, the triangles are congruent by ASA.

Yes. After converting units and finding the missing angles, both triangles have a \(42\,\text{mm}\) side included between angles of \(100^\circ\) and \(35^\circ\), so they are congruent by ASA.

- Apply the triangle inequality to both triangles formed by the diagonal. - Equality in the triangle inequality gives collinear points, not a triangle. - For three given side lengths, use intersections of circles. - To make a convex quadrilateral, place the two remaining vertices on opposite sides of the diagonal.

1. For \(\triangle ABC\), the side lengths would be \(7\,\text{cm}\), \(5\,\text{cm}\), and \(10\,\text{cm}\). Since \(7 + 5 > 10\), this triangle can be formed. 2. For \(\triangle ADC\), the side lengths would be \(4\,\text{cm}\), \(6\,\text{cm}\), and \(10\,\text{cm}\). Since \(4 + 6 = 10\), the points would be collinear and no nondegenerate triangle would form. Therefore, a \(10\,\text{cm}\) diagonal is not possible for the quadrilateral. 3. For \(AC = 8\,\text{cm}\), draw \(\overline{AC}\). Locate \(B\) at an intersection of a circle centered at \(A\) with radius \(7\,\text{cm}\) and a circle centered at \(C\) with radius \(5\,\text{cm}\). 4. On the opposite side of \(\overline{AC}\), locate \(D\) at an intersection of a circle centered at \(A\) with radius \(6\,\text{cm}\) and a circle centered at \(C\) with radius \(4\,\text{cm}\). Connect the vertices in order. 5. Each component triangle is determined by three side lengths, so SSS applies.

a) No. The lengths \(4\,\text{cm}\), \(6\,\text{cm}\), and \(10\,\text{cm}\) form a degenerate triangle because \(4 + 6 = 10\). b) Construct \(\triangle ABC\) and \(\triangle ADC\) on opposite sides of the \(8\,\text{cm}\) diagonal using intersecting circles. Each triangle is determined by SSS.

- A compass circle can represent all points at a fixed distance from one endpoint of the diagonal. - Use the known diagonal as the common baseline for locating both remaining vertices. - Think about how two circle intersections lie relative to the line through their centers.

1. Draw \(\overline{AC}\) with length \(7\,\text{cm}\). 2. Draw a circle centered at \(A\) with radius \(6\,\text{cm}\) and a circle centered at \(C\) with radius \(5\,\text{cm}\). Choose either intersection as \(B\). 3. Draw a circle centered at \(A\) with radius \(3\,\text{cm}\) and a circle centered at \(C\) with radius \(5\,\text{cm}\). 4. Since \(|5-3|<7<5+3\), the second pair of circles has two intersections, one on each side of \(\overline{AC}\). These are the two possible locations for \(D\). 5. Once \(B\) is fixed, choose \(D\) on the opposite side of \(\overline{AC}\) from \(B\) to obtain the convex quadrilateral. The other location produces the concave configuration.

Construct \(B\) as an intersection of circles centered at \(A\) and \(C\) with radii \(6\,\text{cm}\) and \(5\,\text{cm}\). Construct \(D\) as an intersection of circles centered at \(A\) and \(C\) with radii \(3\,\text{cm}\) and \(5\,\text{cm}\). After \(A\), \(B\), and \(C\) are fixed, there are two possible locations for \(D\); the location on the opposite side of \(AC\) from \(B\) gives the convex quadrilateral.

- Which information is already implied by the two side lengths? - When do two sides and an included angle determine a triangle? - Can the two sides and the diagonal form a triangle, and which criterion applies?

1. For a), the side lengths already determine the perimeter: \(P = 2(a+b) = 2(5+3) = 16\,\text{cm}\). This condition gives no angle information, so it is not sufficient. 2. For b), the two adjacent sides and their included angle determine \(\triangle ABD\) by SAS. The fourth vertex of the parallelogram is then fixed, so the condition is sufficient. 3. For c), \(\triangle ABD\) has side lengths \(5\,\text{cm}\), \(3\,\text{cm}\), and \(7\,\text{cm}\). Since \(5 + 3 > 7\), the triangle exists and is determined by SSS. The parallelogram is then determined up to reflection, so the condition is sufficient.

a) Not sufficient; the perimeter is already determined by the side lengths. b) Sufficient; the parallelogram is determined using SAS. c) Sufficient; the diagonal and two adjacent sides determine a triangle by SSS.

- Build the figure one vertex at a time rather than trying to place all four vertices at once. - A copied angle determines a ray once the side of the previous segment is chosen. - A copied segment length determines a unique point on a chosen ray.

1. Draw \(\overline{AB}\) with the given length. 2. At \(B\), copy the given angle \(\angle B\). On the chosen ray, mark \(C\) so that \(BC\) has the given length. 3. At \(C\), copy the given angle \(\angle C\) on the side of \(\overline{BC}\) required by the stated convex orientation. 4. On that ray, mark \(D\) so that \(CD\) has the given length, then connect \(D\) to \(A\). 5. Each copied angle fixes a ray once its side of the preceding segment is chosen, and each copied length fixes one point on that ray. The convex-orientation condition removes the reflected alternative, so the resulting quadrilateral is unique up to congruence.

a) Copy \(AB\), then copy \(\angle B\), copy \(BC\) along that ray, copy \(\angle C\) on the side required by the convex orientation, and copy \(CD\) along the new ray before joining \(D\) to \(A\). b) Each angle fixes a ray and each following length fixes one point on that ray. Specifying the convex orientation removes the mirror-image choice, so the quadrilateral is unique up to congruence.

- What is true about opposite sides and opposite angles of a parallelogram? - How are adjacent angle measures related? - Identify the included angle between each pair of corresponding sides.

1. Opposite angles of a parallelogram are congruent, so \(\angle C = \angle A = 60^\circ\). 2. Adjacent angles of a parallelogram are supplementary, so \(\angle B = \angle D = 180^\circ - 60^\circ = 120^\circ\). 3. Opposite sides of a parallelogram are congruent: \(AB = CD = 5\,\text{cm}\) and \(BC = DA = 3\,\text{cm}\). 4. The equal \(120^\circ\) angles are included between these corresponding side pairs. Therefore, \(\triangle ABC \cong \triangle CDA\) by SAS.

a) \(\angle C = 60^\circ\) b) \(AB = CD\), \(BC = DA\), and \(\angle B = \angle D = 120^\circ\), so the triangles are congruent by SAS.

- Two equal angle measures make triangles similar, but what fixes their scale? - Is a side between the two angles the same corresponding side as a side opposite one of them? - Recall what the order of the letters in ASA and AAS describes.

1. Two angle measures determine the third angle, so any triangles with those angle measures are similar. 2. A side length fixes the scale only when the side has the same corresponding position in both triangles. 3. For example, in one triangle the given side could lie between the two stated angles, while in the other triangle the equal-length side could be opposite one of those angles. 4. Those equal-length sides would not be corresponding sides, so the triangles can have different sizes. Lucas's claim is false as stated. 5. The triangles are congruent by ASA or AAS only when the equal side is in the same corresponding position relative to the equal angles.

No. Two angles and one side determine a unique triangle only when the side corresponds to the same side in both triangles. With the proper correspondence, ASA or AAS proves congruence.

- 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 the triangle inequality when three side lengths are given. - Ask whether angle measures alone determine a triangle's size. - For c), determine whether the given angle is included between the two given sides.

1. For a), the triangle inequality fails because \(5 + 3 = 8 < 9\). No triangle exists. 2. For b), three angles determine the shape but not the size. Infinitely many similar, noncongruent triangles have these angle measures. 3. For c), sides \(a\) and \(c\) meet at angle \(\beta\). Two sides and their included angle determine exactly one triangle by SAS.

a) No triangle exists because \(5 + 3 < 9\). b) The triangle is not uniquely determined; AAA determines similarity, not congruence. c) Exactly one triangle is determined by SAS.

- Identify the side lengths shared by both triangles. - Identify the angle at \(A\). - Decide whether the angle is included between the two known sides. - Recall why SSA is ambiguous.

1. Both triangles have \(AB = 8\,\text{mi}\), \(BC_1 = BC_2 = 5\,\text{mi}\), and the same \(30^\circ\) angle at \(A\). 2. These data give two sides and a nonincluded angle. This is the ambiguous SSA case, which can produce two different triangles. 3. Therefore, \(ABC_1\) and \(ABC_2\) are not congruent.

a) \(AB = 8\,\text{mi}\), \(BC_1 = BC_2 = 5\,\text{mi}\), and \(\angle A = 30^\circ\). b) The data form the ambiguous SSA case, which does not determine a unique triangle.

- 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.

- Divide the quadrilateral into two triangles. - Identify three measurements that determine the first triangle. - Decide which two additional measurements locate the fourth vertex. - Use one of the standard triangle congruence criteria.

1. One valid choice is \(a\), \(b\), \(c\), \(d\), and \(e\). 2. Construct \(\triangle ABC\) from side lengths \(a\), \(b\), and \(e\). It is unique up to reflection by SSS. 3. On the side of \(\overline{AC}\) opposite \(B\), construct \(\triangle ACD\) from side lengths \(c\), \(d\), and \(e\). It is also unique by SSS. 4. Therefore, the convex quadrilateral \(ABCD\) is unique up to reflection.

a) One valid choice is \(a\), \(b\), \(c\), \(d\), and \(e\). b) Construct \(\triangle ABC\) from \(a\), \(b\), and \(e\), then construct \(\triangle ACD\) on the opposite side of \(\overline{AC}\) from \(c\), \(d\), and \(e\). c) Both triangles are determined by SSS, so the convex quadrilateral is unique up to reflection.

- 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 \(AB=CB\).

- 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 \(\angle DAC\).

- 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.

- First construct the triangle containing the given included angle. - How is the remaining vertex located using its distances from \(B\) and \(D\)? - What do zero, one, or two circle intersections mean geometrically? - Does a tangent intersection form a nondegenerate quadrilateral?

1. Sides \(a\) and \(d\) with included angle \(\alpha\) determine \(\triangle ABD\) by SAS. Therefore, vertices \(A\), \(B\), and \(D\), as well as diagonal \(BD\), are fixed up to congruence. 2. Point \(C\) must lie on a circle centered at \(B\) with radius \(b\) and on a circle centered at \(D\) with radius \(c\). 3. The circles may have no intersection, so no quadrilateral exists. If they have two intersections, they generally produce two different placements of \(C\), such as a convex placement and a concave or self-intersecting placement. 4. If the circles are tangent, their single intersection lies on line \(BD\), producing a degenerate figure rather than a proper quadrilateral. 5. Therefore, the five measurements do not always determine exactly one nondegenerate quadrilateral.

Lucas is incorrect. The first component triangle is fixed by SAS, but the two circles used to locate \(C\) may have no intersection or two intersections. A single tangent intersection is degenerate, so the data do not always determine exactly one proper quadrilateral.

- Begin with the longer base. - Use the given angle and leg length to locate \(D\). - How does \(CD \parallel AB\) locate \(C\)? - In the parallelogram decomposition, find \(AE\) from the two base lengths.

1. Draw \(\overline{AB}\) with length \(7\,\text{cm}\). At \(A\), construct a \(70^\circ\) angle and mark \(D\) on its ray so that \(AD = 4\,\text{cm}\). 2. Through \(D\), draw a line parallel to \(AB\). On that line, in the same direction as \(A\) to \(B\), mark \(C\) so that \(DC = 3\,\text{cm}\). Connect \(B\) to \(C\). 3. If the line through \(D\) parallel to \(BC\) meets \(AB\) at \(E\), then \(EBCD\) is a parallelogram. Thus, \(EB = CD = 3\,\text{cm}\). 4. Therefore, \(AE = AB - EB = 7\,\text{cm} - 3\,\text{cm} = 4\,\text{cm}\). 5. In \(\triangle AED\), \(AE = 4\,\text{cm}\), \(AD = 4\,\text{cm}\), and the included angle is \(70^\circ\). The triangle is uniquely determined by SAS.

a) Draw \(AB\), construct the \(70^\circ\) angle at \(A\), mark \(AD = 4\,\text{cm}\), draw the parallel through \(D\), mark \(DC = 3\,\text{cm}\), and connect \(B\) to \(C\). b) SAS. The component triangle has \(AE = AD = 4\,\text{cm}\) with included angle \(70^\circ\).

- 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 \(AC\).

- Check whether the angle is included between the two given sides. - Compare each data set with SSS, SAS, ASA, and AAS. - Remember that SSA is not a general triangle congruence criterion.

1. In part a, one side and its two adjacent angles are given. The triangle is uniquely determined by ASA. 2. In part b, two sides and a nonincluded angle are given. This is the ambiguous SSA case, so the data can produce two noncongruent triangles and do not determine a unique triangle. 3. In part c, two sides and the included angle are given. The triangle is uniquely determined by SAS. 4. In part d, all three side lengths are given. The triangle is uniquely determined by SSS.

a) Yes, by ASA. b) No. SSA is ambiguous here and does not determine a unique triangle. c) Yes, by SAS. d) Yes, by SSS.

- 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.