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Transcript
Aim: How to prove triangles are congruent using a 3rd shortcut: ASA. Do Now: Given: T is the midpoint of PQ, PQ bisects RS, and RQ SP. Explain how RTQ STP. S P T R Q Aim: Triangle Congruence - ASA Course: Applied Geometry Do Now You are given: T is the midpoint of PQ, PQ bisects RS, and RQ SP. Explain how RTQ STP. S P RQ SP – we’re told so T R Q (S S) PT TQ – a midpoint of a segment cuts the segment into (S S) two congruent parts RT TS – a bisector divides a segment into 2 congruent parts (S S) RTQ STP because of SSS SSS Aim: Triangle Congruence - ASA Course: Applied Geometry Sketch 14 – Shortcut #3 A B C Copied 2 angles and included side: BC B’C’, B B’, C C’ A‘ B’ C’ Measurements ABC showed: Shortcut for proving congruence inCongruence - ASA Aim: Triangle triangles: A’B’C’ ASA ASACourse: Applied Geometry Angle-Side-Angle III. ASA = ASA Two triangles are congruent if two angles and the included side of one triangle are equal in measure to two angles and the included side of the other triangle. A’ A B C B’ C’ If A = A', AB = A'B', B = B', then ABC = A'B'C' If ASA ASA , then the triangles are congruent Aim: Triangle Congruence - ASA Course: Applied Geometry Model Problems Is the given information sufficient to prove congruent triangles? C YES C B A D YES E A NO Aim: Triangle Congruence - ASA D Course: Applied Geometry B Model Problems Name the pair of corresponding sides that would have to be proved congruent in order to prove that the triangles are congruent by ASA. DCA CAB C D B D DFA BFC F A B D A C C DB DB A Aim: Triangle Congruence - ASA B Course: Applied Geometry Model Problem CD and AB are straight lines which intersect at E. BA bisects CD. AC CD, BD CD. Explain how ACE BDE using ASA B C D E A C D – lines form right angles and all right angles are & equal 90o (A A) CE ED – bisector cuts segment into 2 parts (S S) CEA BED – intersecting straight lines (A A) form vertical angles which are opposite and ACE BDE because of ASA ASA Aim: Triangle Congruence - ASA Course: Applied Geometry Model Problem 1 2, D is midpoint of EC, 3 4. Explain how AED BCD using ASA A 1 E B 3 4 2 C D 1 2 – Given: we’re told so (A A) ED DC – a midpoint of a segment cuts the segment into two congruent parts (S S) 3 4 – Given: we’re told so (A A) AED BCD because of ASA ASA Aim: Triangle Congruence - ASA Course: Applied Geometry Model Problem DA is a straight line, E B, ED AB, FD DE, CA AB Explain how DEF ABC using ASA E D C F A B E B – Given: we’re told so (A A) ED AB – Given: we’re told so (S S) EDF BAC - lines form right angles and all right angles are & equal 90o (A A) DEF Aim: ABC because of ASA ASA Triangle Congruence - ASA Course: Applied Geometry Aim: Triangle Congruence - ASA Course: Applied Geometry