Triangle Right-Angle-Hypotenuse-Side Equality

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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.


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:

$BC = \sqrt {AB^2 + AC^2}$


$EF = \sqrt {DE^2 + DF^2}$
$\therefore BC = \sqrt {AB^2 + AC^2} = \sqrt {DE^2 + DF^2} = EF$

The part that the remaining two angles are equal to their respective remaining angles follows from Triangle Side-Side-Side Equality.