Math Core

Draw one segment inside a right triangle, the altitude from the right angle to the hypotenuse, and you suddenly have three similar triangles. Their proportions give short formulas for lengths that would otherwise be hard to find, and they even lead to a proof of the Pythagorean theorem.

Three similar triangles

Let $\triangle ABC$ have a right angle at $C$ . The hypotenuse is $\overline{AB}$ . Draw the altitude $\overline{CD}$ from $C$ perpendicular to $\overline{AB}$ . It splits $\triangle ABC$ into two smaller right triangles, $\triangle ACD$ and $\triangle CBD$ .

Right triangle ABC with right angle at C and altitude CD to the hypotenuse.

Each small triangle shares an acute angle with the big one:

$\triangle ACD$ and $\triangle ABC$ both have a right angle ( $\angle ADC$ and $\angle ACB$ ) and share $\angle A$ . By AA, they're similar.
$\triangle CBD$ and $\triangle ABC$ both have a right angle ( $\angle CDB$ and $\angle ACB$ ) and share $\angle B$ . By AA, they're similar.

Since both small triangles are similar to the big one, they're similar to each other too.

Right Triangle Similarity Theorem

If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other:

$\triangle ABC \sim \triangle ACD \sim \triangle CBD.$

The order of the letters matters. In each name, the first letter is the vertex with the angle equal to $\angle A$ , the second has the angle equal to $\angle B$ , and the third is the right angle.

angle	in $\triangle ABC$	in $\triangle ACD$	in $\triangle CBD$
equal to $\angle A$	$A$	$A$	$C$
equal to $\angle B$	$B$	$C$	$B$
right angle	$C$	$D$	$D$

Why is the angle at $C$ in $\triangle CBD$ equal to $\angle A$ ? In $\triangle CBD$ , $m\angle BCD = 90^\circ - m\angle B$ , and in $\triangle ABC$ , $m\angle A = 90^\circ - m\angle B$ as well.

Common mistake

It's easy to match the vertices wrong here, because the triangles overlap and are turned in different directions. Match by angles: right angle to right angle, and the angle equal to $\angle A$ to $\angle A$ . Use the table rather than the look of the picture.

The geometric mean

The proportions that come out of these triangles all have the same form: $\dfrac{a}{x} = \dfrac{x}{b}$ , with the same unknown in two places.

Definition

Geometric mean

The geometric mean of two positive numbers $a$ and $b$ is the positive number $x$ with $\dfrac{a}{x} = \dfrac{x}{b}$ . Cross-multiplying gives $x^2 = ab$ , so $x = \sqrt{ab}$ .

Worked example: Computing geometric means

a. The geometric mean of $4$ and $9$ is $\sqrt{4 \cdot 9} = \sqrt{36} = 6$ . Check: $\dfrac{4}{6} = \dfrac{6}{9}$ .

b. The geometric mean of $2$ and $10$ is $\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt5 \approx 4.47$ .

Two geometric mean theorems

Label the pieces of the hypotenuse: $AD$ is the piece next to leg $\overline{AC}$ , and $DB$ is the piece next to leg $\overline{BC}$ .

The altitude. In $\triangle ACD \sim \triangle CBD$ , the vertices match as $A \leftrightarrow C$ , $C \leftrightarrow B$ , $D \leftrightarrow D$ . So leg $\overline{AD}$ matches leg $\overline{CD}$ , and leg $\overline{CD}$ matches leg $\overline{DB}$ : $\dfrac{AD}{CD} = \dfrac{CD}{DB}$ . So the altitude is the geometric mean of the two pieces of the hypotenuse.

A leg. In $\triangle ABC \sim \triangle ACD$ , the vertices match as $A \leftrightarrow A$ , $B \leftrightarrow C$ , $C \leftrightarrow D$ . So $\overline{AB}$ matches $\overline{AC}$ , and $\overline{AC}$ matches $\overline{AD}$ : $\dfrac{AB}{AC} = \dfrac{AC}{AD}$ . So each leg is the geometric mean of the whole hypotenuse and the piece next to that leg. Similarly $\dfrac{AB}{BC} = \dfrac{BC}{DB}$ .

Geometric mean theorems

In right triangle $ABC$ with altitude $\overline{CD}$ to the hypotenuse:

Altitude rule: $CD^2 = AD \cdot DB$ .
Leg rule: $AC^2 = AD \cdot AB$ and $BC^2 = DB \cdot AB$ .

A way to remember the leg rule: each leg is paired with the piece of the hypotenuse touching that leg, times the whole hypotenuse.

Worked example: Using the altitude rule

In the figure, $AD = 4$ and $DB = 9$ . Find $CD$ .

$CD^2 = AD \cdot DB = 4 \cdot 9 = 36, \qquad CD = 6.$

Worked example: Using the leg rule

In the figure, $AD = 3$ and $AB = 12$ . Find $AC$ and $BC$ .

$AC^2 = AD \cdot AB = 3 \cdot 12 = 36$ , so $AC = 6$ . The other piece is $DB = 12 - 3 = 9$ , so $BC^2 = DB \cdot AB = 9 \cdot 12 = 108$ and $BC = \sqrt{108} = 6\sqrt3$ .

Worked example: Working backward

In the figure, $CD = 6$ and $AD = 4$ . Find $DB$ , $AB$ and $AC$ .

From the altitude rule, $6^2 = 4 \cdot DB$ , so $DB = 9$ . Then $AB = 4 + 9 = 13$ . From the leg rule, $AC^2 = AD \cdot AB = 4 \cdot 13 = 52$ , so $AC = \sqrt{52} = 2\sqrt{13}$ .

A proof of the Pythagorean theorem

The leg rule gives one of the neatest proofs of the Pythagorean theorem, which you'll use heavily in the next unit. Call the legs $a = BC$ and $b = AC$ , the hypotenuse $c = AB$ , and the pieces $p = AD$ and $q = DB$ , so $p + q = c$ . The leg rule says $b^2 = pc$ and $a^2 = qc$ . Add them:

$a^2 + b^2 = qc + pc = (p + q)c = c \cdot c = c^2.$

Tip

When the numbers come out whole, check your answers with the Pythagorean theorem. In the last example, $\triangle ACD$ has legs $4$ and $6$ , so $AC^2$ should be $16 + 36 = 52$ , which matches the leg rule.