Triangle Right-Angle-Hypotenuse-Side Congruence

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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

HL.png

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

Triangle Right-Angle-Hypotenuse-Side Congruence 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

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