The Geometric Mean Theorem

An interactive proof of the geometric mean relations that arise when an altitude is drawn from the right-angle vertex to the hypotenuse of a right triangle, creating three similar triangles.

Keywords: geometric mean, altitude on hypotenuse, similar triangles, right triangle, mean proportional

Prerequisites: triangle-similarity · Difficulty: intermediate

Drop the altitude from the right-angle vertex $C$ to the hypotenuse $A B$ of a right triangle. Its foot $D$ splits the triangle into two smaller ones, and all three triangles turn out similar:

△ A B C \sim △ A C D \sim △ C B D

Strategy: each pair of triangles shares an acute angle and has a right angle, so they are similar by AA. The corresponding-side ratios yield three geometric-mean relations on the hypotenuse segments.

Step 1: The Right Triangle

We start with right triangle $A B C$ where $C$ is the right-angle vertex, $A B$ is the hypotenuse, and $B C$ , $C A$ are the legs.

Step 2: Drop the Altitude to the Hypotenuse

Draw the altitude from $C$ perpendicular to the hypotenuse $A B$ . The foot of the altitude is point $D$ . This splits $A B$ into two segments: $A D$ and $D B$ .

Step 3: Geometric Mean Theorem

Geometric Mean Theorem. When the altitude from the right-angle vertex $C$ meets the hypotenuse $A B$ at $D$ , three pairs of similar triangles arise and three geometric-mean relations hold: $ $C D^{2} = A D \cdot D B, A C^{2} = A D \cdot A B, B C^{2} = B D \cdot B A .$ $

Proof strategy. Each pair of triangles ( $A B C$ vs $A C D$ , $A B C$ vs $C B D$ , $A C D$ vs $C B D$ ) shares an acute angle and has a right angle, so they're similar by AA. The corresponding-side ratios from each pair give the three geometric-mean equations. Adding the last two recovers the Pythagorean theorem.

Step 4: $△ A B C \sim △ A C D$

Triangles $A B C$ and $A C D$ share angle $A$ . Both have a right angle ( $C$ in $△ A B C$ , $D$ in $△ A C D$ ). By AA similarity:

△ A B C \sim △ A C D

From this similarity: $A C / A B = A D / A C$ , giving $A C^{2} = A D \cdot A B$ .

Step 5: $△ A B C \sim △ C B D$

Triangles $A B C$ and $C B D$ share angle $B$ . Both have a right angle ( $C$ in $△ A B C$ , $D$ in $△ C B D$ ). By AA similarity:

△ A B C \sim △ C B D

From this similarity: $B C / B A = B D / B C$ , giving $B C^{2} = B D \cdot B A$ .

Step 6: The Geometric Mean Relations

From $△ A C D \sim △ C B D$ (both similar to $△ A B C$ ), corresponding sides give:

\frac{C D}{D B} = \frac{A D}{C D} ⟹ C D^{2} = A D \cdot D B

Together, the three geometric mean relations are:

$C D^{2} = A D \cdot D B$
$A C^{2} = A D \cdot A B$
$B C^{2} = B D \cdot B A$

Adding the last two: $A C^{2} + B C^{2} = A B (A D + D B) = A B^{2}$ , which is the Pythagorean theorem!

Dropping the altitude from the right-angle vertex to the hypotenuse produces three similar triangles and three geometric mean relations:

$C D^{2} = A D \cdot D B$ (altitude is the geometric mean of the two hypotenuse segments)
$A C^{2} = A D \cdot A B$ (each leg is the geometric mean of its adjacent segment and the whole hypotenuse)
$B C^{2} = B D \cdot B A$

These relations also give a beautiful proof of the Pythagorean theorem: adding the two leg rules yields $A C^{2} + B C^{2} = A B (A D + D B) = A B^{2}$ . ∎

Notes

Three Similar Triangles

All three triangles are similar to each other:

△ A B C \sim △ A C D \sim △ C B D

This follows by AA similarity: each pair shares an acute angle and both have a right angle.

The Geometric Mean Relations

From the similarity ratios we derive three important relations:

Altitude rule: $C D^{2} = A D \cdot D B$
Leg rule (leg $b$ ): $A C^{2} = A D \cdot A B$
Leg rule (leg $a$ ): $B C^{2} = B D \cdot B A$

Each quantity is the geometric mean of two segments of the hypotenuse (or of a segment and the whole hypotenuse).