Side-Angle-Side Congruence by basic rigid motions A geometric realization of a proof in H. Wu’s “Teaching Geometry According to the Common Core Standards”

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Presentation transcript:

Side-Angle-Side Congruence by basic rigid motions A geometric realization of a proof in H. Wu’s “Teaching Geometry According to the Common Core Standards”

Given two triangles, ABC and A 0 B 0 C 0. Assume two pairs of equal corresponding sides with the angle between them equal. We want to prove the triangles are congruent. A B C A0A0 C0C0 B0B0 side angle side

angle In other words, given  ABC and  A 0 B 0 C 0, A B C A0A0 C0C0 B0B0  A =  A 0, |AB| = |A 0 B 0 |, and |AC| = |A 0 C 0 |, we must give a composition of basic rigid motions that maps  ABC exactly onto A0B0C0.A0B0C0. side with

A B C A0A0 C0C0 B0B0

A B C A0A0 C0C0 B0B0

A B C A0A0 C0C0 B0B0

Now we have two of the red triangle’s vertices coinciding with A 0 and B 0 of  A 0 B 0 C 0. A B C A0A0 C0C0 B0B0 After a reflection of the red triangle across A 0 B 0, the third vertex will exactly coincide with C 0.

Can we be sure this composition of basic rigid motions A B C A0A0 C0C0 B0B0 takes C to C 0 — and the red triangle with it? (the reflection of the rotation of the translation of the image of  ABC)

Yes! The two marked angles at A 0 are equal since basic rigid motions preserve degrees of angles, A B C A0A0 C0C0 B0B0 and  CAB =  C 0 A 0 B 0 is given by hypothesis. A reflection across A 0 B 0 does take C to C 0 — and the red triangle with it!

A B C A0A0 C0C0 B0B0 Since basic rigid motions preserve length and since |AC| = |A 0 C 0 |, by Lemma 8, the red triangle coincides with  A 0 B 0 C 0. after a reflection across A 0 B 0, The triangles are congruent. Our proof is complete.

Given two triangles with two pairs of equal sides and an included equal angle, maps the image of one triangle onto the other. Therefore, the triangles are congruent. basic rigid motions A B C A0A0 C0C0 B0B0 A0A0 C0C0 B0B0 a composition of (translation,rotation, and reflection)