Triangle Right-Angle-Hypotenuse-Side Congruence/Proof 2

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

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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:

$\triangle ABD \cong \triangle ACD$

$\blacksquare$