"Given two triangles, [math]\triangle ABC[/math] and [math]\triangle DEF[/math] with right angles at [math]\angle A[/math] and [math]\angle D[/math], [math]\overline{AB}\cong\overline{DE}[/math], and [math]\overline{BC}\cong\overline{EF}[/math]. Prove that [math]\triangle ABC\cong\triangle DEF[/math]."[br][br][b]Proof:[/b][br]Consider the point X, on the line [math]\overline{AC}[/math] such that X is on the opposite side of A from C and such that [math]\overline{AX}\cong\overline{DF}[/math]. By Euclid's Proposition 13, [math]\angle BAX[/math] is a right angle. Thus, [math]\angle BAX\cong\angle EDF[/math]. We also know that [math]\overline{AB}\cong\overline{DE}[/math]. Therefore, we have satisfied the SAS criterion for congruent triangles. That is, [math]\triangle ABX\cong\triangle DEF[/math]. [br][br]Notice then, that [math]\overline{BX}\cong\overline{EF}\cong\overline{BC}[/math]. Thus, [math]\triangle BCX[/math] is an isosceles triangle. Therefore, [math]\angle C\cong\angle X\cong\angle F.[/math] Then, the AAS criterion for congruent triangles is satisfied for [math]\triangle ABC[/math] and [math]\triangle DEF[/math]. So, we can conclude that [math]\triangle ABC\cong\triangle DEF[/math].