Triangle Right-Angle-Hypotenuse-Side Congruence/Proof 1
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Theorem
If two right triangles have:
- their hypotenuses equal
- another of their respective sides equal
they will also have:
Proof
Let $\triangle ABC$ and $\triangle DEF$ be two triangles having sides $AB = DE$ and $AC = DF$, and with $\angle ABC = \angle DEF = 90^\circ$.
- $BC = \sqrt {AB^2 + AC^2}$
and:
- $EF = \sqrt {DE^2 + DF^2}$
- $\therefore BC = \sqrt {AB^2 + AC^2} = \sqrt {DE^2 + DF^2} = EF$
The part that the remaining two angles are equal to their respective remaining angles follows from Triangle Side-Side-Side Congruence.
$\blacksquare$