Triangle Right-Angle-Hypotenuse-Side Equality

Theorem
If two triangles have:
 * one right angle each;
 * the sides opposite to the right angle equal;
 * another two respective sides equal

they will also have:
 * their third sides equal;
 * the remaining two angles equal to their respective remaining angles.

Proof
Let $\triangle ABC$ and $\triangle DEF$ be two triangles having sides $AB = DE$ and $AC = DF$, and with $\angle ABC = \angle DEF = 90 ^\circ$.

By Pythagoras' Theorem, $BC = \sqrt{ AB^2 + AC^2 }$ 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 Equality.