The Proof of the Hypotenuse Leg Theorem

Posted on 7 July, 2013 by Math Proofs 1 Comment

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, $ABC \cong DEF$ .

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