# Triangle Right-Angle-Hypotenuse-Side Equality

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## 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 $\angle ABC = \angle DEF = 90^\circ$.

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

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 $1.15$