# The Proof of the Hypotenuse Leg Theorem

The Hypotenuse Leg Theorem states that if the hypotenuse and one leg of a triangle are congruent to the hypotenuse and leg of another triangle, then the two triangles are congruent.

In the figure above, $ABC$ and $DEF$ are right triangles with right angles at $C$ and $F$ and with $AB \cong DE$ and $BE \cong EF$. We are going to show that the two triangles are congruent.

Given

Triangle $ABC$ and $DEF$, right angled at $C$ and $F$ respectively, with $AB \cong DE$ and $BC \cong EF$.

Proof

It is given that $AB \cong DE$, $BC \cong EF$, so by the definition of congruence,

$AB = DE$ and $BC = EF$.

Now, by Pythagorean Theorem,

$AB^2 = AC^2 + BC^2$ and $DE^2 = DF^2 + EF^2$.

Since $DE = AB$, by substitution, we have

$AC^2 + BC^2 = DF^2 + EF^2$

Now, since $BC = EF$,

$AC^2 + EF^2 = DF^2 + EF^2$

Subtracting EF^2 from both sides, we have

$AC = DF$.

So, by the SSS Congruence, $ABC \cong DEF$.