# The Proof of the Right Triangle Altitude Theorem

The geometric mean of two positive integers $a$ and $b$ is $\sqrt{ab}$. In this post, we are going to show the relationship between geometric mean and the relationships among the sides and altitude of a right triangle.

##### The Right Triangle Altitude Theorem

In a right triangle,

• the altitude to the hypotenuse is the geometric mean of the segments into which it divides (the hypotenuse); and
• each leg is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to the leg.

The two theorems above, states that in the triangle below,

$h = \sqrt{xy}$, $a = \sqrt{cx}$ and $b = \sqrt{cy}$.

##### Proof

Let $a$ and $b$ be the legs of the triangle $ABC$ with hypotenuse $c$. Let $h$ be the altitude of the triangle to the hypotenuse dividing the $c$ to segments $x$ and $y$.

First, we show that that the three triangles, $\triangle ABC$, $\triangle ACD$, and $\triangle CBD$ are similar triangles.

$m \angle A = m \angle BCD$ since both of them equal $90 - m \angle B$. So,  $\triangle ABC$, $\triangle ACD$, and $\triangle CBD$ are right triangles which all include  an angle whose measure equal $m \angle A$. Therefore, these three triangles are similar. Since these three triangles are similar, their sides are proportional.

It follows that for $\triangle ACD$ and $\triangle CBD$,

$\displaystyle \frac{x}{h} =\frac{h}{y}$

which implies that

$h^2 = xy$ and $h = \sqrt{xy}$.

It remains to show that $a = \sqrt{cx}$ and $b = \sqrt{cy}$.

For $\triangle ABC$ and $\triangle CBD$, we have

$\displaystyle \frac{x}{a} = \frac{a}{c}$

which means that

$a^2 = xc$ and $a = \sqrt{xc}$.

Also, for $\triangle ABC$ and $\triangle ACD$, we have

$\displaystyle \frac{b}{c} = \frac{y}{b}$

which means that $b^2 = cy$ and that $b = \sqrt{cy}$.

These prove the Right Triangle Altitude Theorem..

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