# Week 8 Day 1 Opener

- Author:
- Katie Akesson

Proof:
1. Segments

*AB*and*DE*are the same length so they are congruent. Therefore, there is a rigid motion that takes*AB*to*DE*. 2. Apply that rigid motion to triangle*ABC*. The image of*A*will coincide with*D*, and the image of*B*will coincide with*E*. 3. We cannot be sure that the image of*C*coincides with*F*yet. If necessary, reflect the image of triangle*ABC*across*DE*to be sure the image of*C*, which we will call*C'*, is on the same side of*DE*as*F*. (This reflection does not change the image of*A*or*B*.) 4. We know the image of angle*A*is congruent to angle*D*because rigid motions don’t change the size of angles. 5.*C'*must be on ray*DF*since both*C'*and*F*are on the same side of*DE*, and make the same angle with it at*D*. 6. Segment*DC'*is the image of*AC*and rigid motions preserve distance, so they must have the same length. 7. We also know*AC*has the same length as*DF*. So*DC'*and*DF*must be the same length. 8. Since*C'*and*F*are the same distance along the same ray from*D*, they have to be in the same place. 9. We have shown that a rigid motion takes*A*to*D*,*B*to*E*, and*C*to*F*; therefore, triangle*ABC*is congruent to triangle*DEF*.