Given two triangles, [math]\Delta[/math]ABC and [math]\Delta[/math]DEF, with right angles [math]\angle[/math]A and [math]\angle[/math]D, AB [math]\cong[/math] DE, and BC [math]\cong[/math] EF. Prove that [math]\Delta[/math]ABC [math]\cong[/math] [math]\Delta[/math]DEF. This is the Hypotenuse-Leg criterion for congruence of right triangles. [br][br]Note: If AC [math]\cong[/math] DF, then these triangles would be congruent because of SAS:[br](we would have AB [math]\cong[/math] DE, [math]\angle[/math]A [math]\cong[/math] [math]\angle[/math]D, and AC [math]\cong[/math] DF)[br][br]Proof: Assume AC is not congruent to DF. By (III-1), there is a point X on ray DF such that AC [math]\cong[/math] DX. [br]By our assumption, we know that X is different from F. We know that [math]\Delta[/math]ABC [math]\cong[/math] [math]\Delta[/math]DEX because [br]AB [math]\cong[/math] DE, [math]\angle[/math]A [math]\cong[/math] [math]\angle[/math]D, and AC [math]\cong[/math] DX. This implies that BC [math]\cong[/math] EX but this is not possible because [br]BC [math]\cong[/math] EF by our assumption. Thus, X must be point F. Therefore, [math]\Delta[/math]ABC [math]\cong[/math] [math]\Delta[/math]DEF.