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Aim: Triangle Congruence - SSS Course: Applied Geometry Do Now: Aim: How to prove triangles are congruent using a 2 nd shortcut: SSS.

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Presentation on theme: "Aim: Triangle Congruence - SSS Course: Applied Geometry Do Now: Aim: How to prove triangles are congruent using a 2 nd shortcut: SSS."— Presentation transcript:

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2 Aim: Triangle Congruence - SSS Course: Applied Geometry Do Now: Aim: How to prove triangles are congruent using a 2 nd shortcut: SSS.

3 Aim: Triangle Congruence - SSS Course: Applied Geometry Sketch 13 – Shortcut #2 SSS  SSS Copied 3 sides AB  A’B’, BC  B’C’, BC  B’C’ Copied 3 sides AB  A’B’, BC  B’C’, BC  B’C’ Shortcut for proving congruence in triangles: Measurements showed:  ABC   A’B’C’ B A C B’ A’ C’

4 Aim: Triangle Congruence - SSS Course: Applied Geometry Side-Side-Side II. SSS = SSS Two triangles are congruent if the three sides of one triangle are equal in measure to the three sides of the other triangle. S represents a side of the triangle. A BB’CC’ A’ If AC = A'C', CB = C'B', BA = B'A', then  ABC =  A'B'C' If SSS  SSS, then the triangles are congruent.

5 Aim: Triangle Congruence - SSS Course: Applied Geometry Model Problems Is the given information sufficient to prove congruent triangles?

6 Aim: Triangle Congruence - SSS Course: Applied Geometry 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 SSS.

7 Aim: Triangle Congruence - SSS Course: Applied Geometry Model Problem You are given: Isosceles triangle ABC with CA  CB with D the midpoint of base AB. Explain how  ACD   BCD CA  CB – we’re told so AD  DB – a midpoint of a segment cuts the segment into two congruent parts CD  CD – any figure is equal to itself The two triangles are congruent because of SSS  SSS (S  S)

8 Aim: Triangle Congruence - SSS Course: Applied Geometry Model Problem You are given: T is the midpoint of PQ, PQ bisects RS, and RQ  SP. Explain how  RTQ   STP. RQ  SP – we’re told so PT  TQ – a midpoint of a segment cuts the segment into two congruent parts RT  TS – a bisector divides a segment into 2 congruent parts  RTQ   STP because of SSS  SSS (S  S)

9 Aim: Triangle Congruence - SSS Course: Applied Geometry


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