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

When the altitude from the right-angle vertex $C$ is drawn to the hypotenuse $AB$ of a right triangle, it creates point $D$ on $AB$ and splits the triangle into two smaller triangles.

Three Similar Triangles

All three triangles are similar to each other:

\triangle ABC \sim \triangle ACD \sim \triangle CBD
    

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:

1. Altitude rule: $CD^2 = AD \cdot DB$
2. Leg rule (leg $b$): $AC^2 = AD \cdot AB$
3. Leg rule (leg $a$): $BC^2 = BD \cdot BA$

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

The Proof

Step 1: The Right Triangle

We start with right triangle $ABC$ where $C$ is the right-angle vertex, $AB$ is the hypotenuse, and $BC$, $CA$ are the legs.

Step 2: Drop the Altitude to the Hypotenuse

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

Step 3: Geometric Mean Theorem

Geometric Mean Theorem. When the altitude from the right-angle vertex $C$ meets the hypotenuse $AB$ at $D$, three pairs of similar triangles arise and three geometric-mean relations hold: $$CD^2 = AD \cdot DB, \quad AC^2 = AD \cdot AB, \quad BC^2 = BD \cdot BA.$$

Proof strategy. Each pair of triangles ($ABC$ vs $ACD$, $ABC$ vs $CBD$, $ACD$ vs $CBD$) 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: $\triangle ABC \sim \triangle ACD$

Triangles $ABC$ and $ACD$ share angle $A$. Both have a right angle ($C$ in $\triangle ABC$, $D$ in $\triangle ACD$). By AA similarity:

\triangle ABC \sim \triangle ACD
    

From this similarity: $AC/AB = AD/AC$, giving $AC^2 = AD \cdot AB$.

Step 5: $\triangle ABC \sim \triangle CBD$

Triangles $ABC$ and $CBD$ share angle $B$. Both have a right angle ($C$ in $\triangle ABC$, $D$ in $\triangle CBD$). By AA similarity:

\triangle ABC \sim \triangle CBD
    

From this similarity: $BC/BA = BD/BC$, giving $BC^2 = BD \cdot BA$.

Step 6: The Geometric Mean Relations

From $\triangle ACD \sim \triangle CBD$ (both similar to $\triangle ABC$), corresponding sides give:

\frac{CD}{DB} = \frac{AD}{CD} \implies CD^2 = AD \cdot DB
    

Together, the three geometric mean relations are:

• $CD^2 = AD \cdot DB$
• $AC^2 = AD \cdot AB$
• $BC^2 = BD \cdot BA$

Adding the last two: $AC^2 + BC^2 = AB(AD + DB) = AB^2$, which is the Pythagorean theorem!

Conclusion

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

• $CD^2 = AD \cdot DB$ (altitude is the geometric mean of the two hypotenuse segments)
• $AC^2 = AD \cdot AB$ (each leg is the geometric mean of its adjacent segment and the whole hypotenuse)
• $BC^2 = BD \cdot BA$

These relations also give a beautiful proof of the Pythagorean theorem: adding the two leg rules yields $AC^2 + BC^2 = AB(AD + DB) = AB^2$.