AMSCO'S Geometry. New York - Rye High School

216 Transformations and the Coordinate Plane(1) Let the image of A be A and the image ofB be B under reflection in k. Let C be themidpoint of AAr and D be the midpoint ofBBr. Points C and D are on k, since k is theperpendicular bisector of AAr and of BBr.(2) BD > BrD since D is the midpoint of BBr,BDC BDC since perpendicular linesintersect to form right angles, andCD > CD by the reflexive property. Thus,BDC BDC by SAS.(3) From step 2, we can conclude thatBBC > BrC and BCD BCD sincecorresponding parts of congruent trianglesA Dare congruent.C(4) Since k is the perpendicular bisector ofAAr, ACD and ACD are right angles.Thus, ACB and BCD are complementarykA Bangles. Similarly, ACB and BCD are complementary angles. Iftwo angles are congruent, their complements are congruent. Since BCDand BCD were shown to be congruent in step 3, their complements arecongruent: ACB ACB.(5) By SAS, ACB ACB. Since corresponding parts of congruent trianglesare congruent, AB > ArBr. Therefore, AB = AB by the definition ofcongruent segments.A similar proof can be given when A and B are on opposite sides of the linek. The proof is left to the student. (See exercise 11.)kACBDBSince distance is preservedunder a line reflection, we can provethat the image of a triangle is a congruenttriangle and that angle measure,collinearity, and midpoint arealso preserved. For each of the followingcorollaries, the image of A isA, the image of B is B, and theimage of C is C. Therefore, in thediagram, ABC ABC by SSS.We will use this fact to prove the followingcorollaries:BCDkMAM