* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Congruence by S.A.S.
Plane of rotation wikipedia , lookup
Euler angles wikipedia , lookup
Rational trigonometry wikipedia , lookup
Four color theorem wikipedia , lookup
Trigonometric functions wikipedia , lookup
Reuleaux triangle wikipedia , lookup
History of trigonometry wikipedia , lookup
Pythagorean theorem wikipedia , lookup
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 A0B0C0. Assume two pairs of equal corresponding sides with the angle between them equal. C B0 side C0 A side B A0 We want to prove the triangles are congruent. In other words, given ABC and A0B0C0, with |AB| = |A0B0|, A = A0, and |AC| = |A0C0|, B0 C side C0 A side B A0 we must give a composition of basic rigid motions that maps ABC exactly onto A0B0C0. We first move vertex A to A0 by a translation 𝑇 along the vector from A to A0 B0 C A C0 B A0 𝑇 translates all points in the plane. Original positions are shown with dashed lines and new positions in red. Then we use a rotation ℛ to bring the horizontal side of the red triangle (which is the translated image of AB by 𝑇) to A0B0. B0 C A C0 B A0 ℛ maps the translated image of AB exactly onto A0B0 because |AB| = |A0B0| and translations B0 preserve length. C C0 A B A0 Now we have two of the red triangle’s vertices coinciding with A0 and B0 of A0B0C0. B0 C A C0 B A0 After a reflection of the red triangle across A0B0, the third vertex will exactly coincide with C0. Can we be sure this composition of basic rigid motions (the reflection of the rotation of the translation of the B0 image of ABC) C A C0 B A0 takes C to C0 — and the red triangle with it? Yes! The two marked angles at A0 are equal since basic rigid motions preserve degrees of angles, B0 and CAB = C0A0B0 is given by hypothesis. C A C0 B A0 A reflection across A0B0 does take C to C0 — and the red triangle with it! Since basic rigid motions preserve length and since |AC| = |A0C0|, B0 after a reflection across A0B0, C A C0 B A0 by Lemma 8, the red triangle coincides with A0B0C0. The triangles are congruent. Our proof is complete. Given two triangles with two pairs of equal sides and an included equal angle, a composition of B0 basic rigid motions (translation, rotation, and C C0 reflection) A B A0 maps the image of one triangle onto the other. Therefore, the triangles are congruent.