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1) Find m<ACB 2) Find m<CLK 3) Find m<CBP 4) Which two angles are the remote interior angles for <MLK? 5) Find m<BCK 6) Find m<MLK A B 70 L K M P 42 C.

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Presentation on theme: "1) Find m<ACB 2) Find m<CLK 3) Find m<CBP 4) Which two angles are the remote interior angles for <MLK? 5) Find m<BCK 6) Find m<MLK A B 70 L K M P 42 C."— Presentation transcript:

1 1) Find m<ACB 2) Find m<CLK 3) Find m<CBP 4) Which two angles are the remote interior angles for <MLK? 5) Find m<BCK 6) Find m<MLK A B 70 L K M P 42 C 70

2 1) Find m<ACB All the angles in a triangle add up to 180. 180 = 70 + 42 + m<ACB 180 = 112 + m<ACB 68 = m<ACB 2) Find m<CLK The third angle theorem states if two angles in a triangle are congruent to two angles in another triangle, then the thirds angles are also congruent. Since both triangles have angles measuring 42 and 70, m<ACB is congruent to m<CLK. M<ACB = 68 = m<CLK A B 70 L K M P 42 C 70

3 3) Find m<CBP <CBP and <ABC are supplementary. 180 = m<CBP + m<ABC 180 = m<CBP + 42 138 = m<CBP 4) Which two angles are the remote interior angles for <MLK? <LKC and <LCK A B 70 L K M P 42 C 70

4 5) Find m<BCK <ACB, <BCK, and <KCL are supplementary. 180 = m<ACB + m<BCK + m<KCL 180 = 68 + m<BCK + 70 180 = 138 + m<BCK 42 = m<BCK 6) Find m<MLK The exterior angle theorem states that the remote interior angles add up to the exterior angle. m<MLK = m<LKC + m<LCK m<MLK = 42 + 70 m<MLK = 112 A B 70 L K M P 42 C 70

5 Exploring Congruent Triangles Proving Triangles Congruent: SSS, SAS, ASA

6 Congruent Triangles- Triangles that are the same size and same shape. CPCTC (Corresponding parts of congruent triangles are congruent) - Two Triangles are congruent if and only if their corresponding parts are congruent. Congruence Transformations- If you slide, rotate, or flip a figure, congruence will not change. Theorem 4-4- Congruence of triangles is reflexive, symmetric, and transitive. * Use to indicate each correspondence.

7 SSS Postulate- If the sides of one triangle are congruence to the sides of a second triangle, then the triangles are congruent. Included Angle- The angle formed by two sides of a triangle. SAS Postulate- If two sides and the included angle of one triangle is congruent to two sides and the included angle of another triangle, then the triangles are congruent.

8 Included Side- the side between two angles in a triangle. ASA Postulate- If two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, the triangles are congruent.

9 Example 1: If Triangle ABC is congruent to Triangle DEF, name the corresponding congruent angles and sides. D A C B F E AB DE BC EF AC DF <A <D <B <E <C <F

10 Example 3: If Triangle BIG is congruent to Triangle DEN, name the corresponding congruent angles and sides. Draw a figure and mark the corresponding parts. BI DE IG EN BG DN <B <D <I <E <G <N N E D GI B

11 A)Triangle TRV is congruent to _________. Triangle DEG B) Triangle XYZ is congruent to __________. Triangle PVZ C) Triangle YZW is congruent to __________. Triangle WXY Example 5: Complete the congruence statement. Y X Z V P G E D VR T WZ Y X

12 Example 1) Determine which postulate can be used to prove the triangles are congruent. If it is not possible to prove that they are congruent, write not possible. Not possible SAS ASA SSS

13 Example 2: Given Triangle PQR with vertices P(3, 4), Q(2, 2), and R(7, 2) and Triangle STU with vertices S(6, -3), T(4, -2), and U(4, -7), show that Triangle PQR is congruent to Triangle STU. Use the distance formula d=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 ) to find the length of each side. PQ d=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 ) d=√((3 – 2) 2 + (4 – 2) 2 ) d=√((1) 2 + (2) 2 ) d=√(1 + 4) d=√(5) QR d=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 ) d=√((2 – 7) 2 + (2 – 2) 2 ) d=√((-5) 2 + (0) 2 ) d=√(25 + 0) d=√(25) d= 5 PR d=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 ) d=√((3 – 7) 2 + (4 – 2) 2 ) d=√((-4) 2 + (2) 2 ) d=√(16 + 4) d=√(20) ST d=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 ) d=√((6 – 4) 2 + (-3 – -2) 2 ) d=√((6 – 4) 2 + (-3 + 2) 2 ) d=√((2) 2 + (-1) 2 ) d=√(4 + 1) d=√(5) TU d=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 ) d=√((4 – 4) 2 + (-2 – -7) 2 ) d=√((4 – 4) 2 + (-2 + 7) 2 ) d=√((0) 2 + (5) 2 ) d=√(0 + 25) d=√(25) d= 5 SU d=√((x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 ) d=√((6 – 4) 2 + (-3 – -7) 2 ) d=√((6 – 4) 2 + (-3 + 7) 2 ) d=√((2) 2 + (4) 2 ) d=√(4 + 16) d=√(20) Since PQ = ST, QR = TU, and PR = SU, the triangles are congruent by SSS.

14 Example 3) Given: <1 and <2 are right angles, line segment ST is congruent to line segment TP. Prove: Triangle STR is congruent to Triangle PTR. 1 2 3 StatementsReasons <1 and <2 are right angles, line segment ST is congruent to line segment TP. <1 is congruent to <2 Triangle STR is congruent to Triangle PTR. Segment TR is congruent to segment TR Given All right angles are congruent SAS Congruence of segments is reflexive 4 6 5 S T P R

15 Example 4) Given: AB is parallel to DC and AB is congruent to DC Prove: Triangle ABD is congruent to triangle CDB C AB StatementsReasons AB is parallel to DC <1 is congruent to _______ BD is congruent to BD AB is congruent to ________ Given Alternate interior angles are congruent Given S A S <2 DC Triangle ABD is congruent to triangle CDB ___ ___ ___ D 2 1 Reflexive Property of Equality


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