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.

$\blacksquare$

Sources

• 1968: M.N. Aref and William Wernick: Problems & Solutions in Euclidean Geometry ... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.15$