Draw the altitude from the right angle of a right triangle to its hypotenuse. It splits the triangle into two smaller right triangles, and all three triangles are similar. You can prove it with Proving AA Similarity, and it leads to a new proof of the theorem from The Pythagorean Theorem.
Why it works
In right $\triangle ABC$ (right angle at $C$) with altitude $\overline{CD}$, call the angle at $A$ $\alpha$ and the angle at $B$ $\beta$, so $\alpha + \beta = 90^\circ$. $\triangle ACD$ has a right angle at $D$ and the angle $\alpha$ at $A$, so its angle at $C$ is $\beta$. $\triangle CBD$ has a right angle at $D$ and $\beta$ at $B$, so its angle at $C$ is $\alpha$. All three triangles have angles $\alpha$, $\beta$, $90^\circ$: they are similar by AA, with $\triangle ACD \sim \triangle CBD \sim \triangle ABC$.
Key idea
The altitude to the hypotenuse makes three similar right triangles. So the altitude is the geometric mean of the hypotenuse's two pieces, $CD^2 = AD \cdot DB$, and each leg is the geometric mean of the hypotenuse and the piece next to it: $AC^2 = AD \cdot AB$ and $BC^2 = DB \cdot AB$.
Worked example: The altitude
The altitude to the hypotenuse splits it into $AD = 4$ and $DB = 9$. Find $CD$.
- From $\triangle ACD \sim \triangle CBD$: $\frac{AD}{CD} = \frac{CD}{DB}$, so $CD^2 = 4 \cdot 9 = 36$.
- $CD = 6$.
Answer: $CD = 6$
Worked example: A leg
With $AD = 4$ and $DB = 9$ again, find $AC$.
- From $\triangle ACD \sim \triangle ABC$: $\frac{AD}{AC} = \frac{AC}{AB}$, so $AC^2 = AD \cdot AB$.
- $AB = 4 + 9 = 13$, so $AC^2 = 52$ and $AC = \sqrt{52} = 2\sqrt{13}$.
- Check with the Pythagorean theorem in $\triangle ACD$: $4^2 + 6^2 = 52$.
Answer: $AC = 2\sqrt{13}$
Worked example: Working backward
$CD = 12$ and $AD = 8$. Find $DB$.
- $CD^2 = AD \cdot DB$: $144 = 8 \cdot DB$.
- $DB = 18$.
Answer: $DB = 18$
The geometric mean of two positive numbers $p$ and $q$ is $\sqrt{pq}$: the number $x$ with $\frac{p}{x} = \frac{x}{q}$. The altitude to the hypotenuse is the geometric mean of the two pieces it creates.
Common mistake
Pairing a leg with the wrong piece of the hypotenuse. Leg $\overline{AC}$ touches $A$, so it goes with $AD$, the piece next to it: $AC^2 = AD \cdot AB$. Check: each leg's partner piece shares an endpoint with it.