Triangle Right-Angle-Hypotenuse-Side Congruence

Theorem

If two right triangles have:

their hypotenuses equal

another of their respective sides equal

they will also have:

their third sides equal

the remaining two angles equal to their respective remaining angles.

Proof 1

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 Congruence.

$\blacksquare$

Proof 2

Let $\triangle ADB$ and $\triangle ADC$ both be right triangles.

Let them have equal hypotenuse and one leg ($AD$) equal.

by hypothesis:

$AB = AC$

$\angle ADB = \angle ADC = \ $ one right angle

$AD$ is shared

So the two triangles can be drawn as shown with $BD$ and $DC$ joined at $D$.

By addition:

$\angle BDA + \angle CDA = \angle BDC = \ $ two right angles

By Two Angles making Two Right Angles make Straight Line:

$BDC$ is one straight line

the points $BDC$ are collinear

So:

$\triangle ABC$ is a triangle.

By definition of isosceles triangle:

$\triangle ABC$ is isosceles.

Since $AD \perp BDC$, by definition of perpendicular bisector:

$BD = DC$

By Triangle Side-Angle-Side Congruence:

$\triangle ABD \cong \triangle ACD$

$\blacksquare$

Also known as

is also known as RHS or the RHS Condition.

However, because RHS is also used as a standard abbreviation for the right hand side of an equation, it is deprecated on $\mathsf{Pr} \infty \mathsf{fWiki}$ on account of ambiguity.

Sources

1968: M.N. Aref and William Wernick: Problems & Solutions in Euclidean Geometry ... (previous) ... (next): Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.15$
1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): congruent: 1. $(4)$
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): congruent: 1. $(4)$