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

Given

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

Proof

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

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

Now, by Pythagorean Theorem,

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

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

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

Now, since \$latex BC = EF\$,

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

Subtracting EF^2 from both sides, we have

\$latex AC = DF\$.

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