AMSCO'S Geometry. New York - Rye High School

(1) Since any line segment may beextended any required length, extendDE to G so that EG > AB. Draw FG.Proving Right Triangles Congruent by Hypotenuse, Leg 363(2) GEF and DEF form a linear pair,and DEF is a right angle. Therefore,GEF is a right angle. We are givenA B D E Gthat B is a right angle. All right angles are congruent, so B GEF.(3) We are also given BC > EF.(4) Therefore, ABC GEF by SAS.We now show that DEF is also congruent to the constructed triangle, GEF:(5) Since corresponding sides of congruenttriangles are congruent,AC > GF. Since we are givenAC > DF, GF > DF by the transitiveproperty of congruence.(6) If two sides of a triangle are congruent,ACB DFE Gthe angles opposite these sides are congruent. In DFG, GF > DF,so D G. Also, DEF GEF since all right angles are congruent.(7) Therefore, DEF GEF by AAS.(8) Therefore, ABC DEF by the transitive property of congruence(steps 4 and 7).This theorem is called the hypotenuse-leg triangle congruence theorem,abbreviated HL. Therefore, from this point on, when the hypotenuse and a legof one right triangle are congruent to the corresponding parts of a second righttriangle, we may say that the triangles are congruent.A corollary of this theorem is the converse of Corollary 9.12b.CFCorollary 9.14aIf a point is equidistant from the sides of an angle, then it lies on the bisectorof the angle.Given ABC, PD ' BA hat D, PF ' BC hat F, andPD PFProve ABP CBPStrategy Use HL to prove PDB PFB.ADPThe proof of this theorem is left to the student.(See exercise 8.)BFC