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Side-Angle-Side Congruence by basic rigid motions

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gitel
11 diapositives · 318 Views · Aug 04, 2014

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. B 0. C.

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  1. 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”

  2. Given two triangles, ABC and A0B0C0. Assume two pairs of equal corresponding sides with the angle between them equal. B0 C C0 side side angle angle A side B side A0 We want to prove the triangles are congruent.

  3. In other words, given ABC and A0B0C0, with A = A0, |AB| = |A0B0|, B0 and |AC| = |A0C0|, C C0 side side angle angle A side B side A0 we must give a composition of basic rigid motions that maps ABC exactly onto A0B0C0.

  4. We first move vertex A to A0 by a translation along the vector from A to A0 B0 C C0 A B A0 translates all points in the plane. Original positions are shown with dashed lines and new positions in red.

  5. Then we use a rotation to bring the horizontal side of the red triangle (which is the translated image of AB by ) to A0B0. B0 C C0 A B A0

  6. maps the translated image of AB exactly onto A0B0because |AB| = |A0B0| and translations preserve length. B0 C C0 A B A0

  7. Now we have two of the red triangle’s vertices coinciding with A0 and B0 of  A0B0C0. B0 C C0 A B A0 After a reflection of the red triangle across A0B0, the third vertex will exactly coincide with C0.

  8. Can we be surethis composition of basic rigid motions (the reflection of the rotation of the translation of the B0 image of ABC) C C0 A B A0 takes C to C0 — and the red triangle with it?

  9. Yes! The two marked angles at A0are equal since basic rigid motions preserve degrees of angles, and CAB =  C0A0B0 is given by hypothesis. B0 C C0 A B A0 A reflection across A0B0 does take C to C0 — and the red triangle with it!

  10. Since basic rigid motions preserve length and since |AC| = |A0C0|, B0 after a reflection across A0B0, C C0 A B A0 by Lemma 8, the red triangle coincides with A0B0C0. The triangles are congruent. Our proof is complete.

  11. Given two triangles with two pairs of equal sides and an included equal angle, a composition of basic rigid motions B0 B0 (translation, rotation, and reflection) C C0 C0 A B A0 A0 maps the image of one triangle onto the other. Therefore, the triangles are congruent.