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.

right triangle altitude theorem

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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