直角三角形の斜辺への高さ

Problem

In a right triangle, an altitude from the right angle meets the hypotenuse at $D$, with $AD = 4$ and $DB = 9$; find the altitude $CD$ and the legs $AC$ and $BC$.

Step 1: Find the altitude $CD$

Using the geometric mean altitude theorem, the altitude satisfies

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

So,

$$CD = 6.$$

Step 2: Find the leg $AC$

First find the full hypotenuse:

$$AB = AD + DB = 4 + 9 = 13.$$

For leg $AC$, the leg theorem gives

$$AC^2 = AD \cdot AB = 4 \cdot 13 = 52,$$

so

$$AC = 2\sqrt{13}.$$

Step 3: Find the leg $BC$

Using the same leg theorem with the other segment,

$$BC^2 = DB \cdot AB = 9 \cdot 13 = 117,$$

so

$$BC = 3\sqrt{13}.$$

Answer

$$CD = 6,\quad AC = 2\sqrt{13},\quad BC = 3\sqrt{13}.$$