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.

hypotenuse leg theorem

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. 


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


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, $latex ABC \cong DEF$.


1 thought on “The Proof of the Hypotenuse Leg Theorem

  1. Pingback: The Proof of the Hypotenuse Angle Theorem

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