The Altitude to the Hypotenuse

In a right triangle with legs $a, b$ and hypotenuse $c$, the altitude $p$ to the hypotenuse satisfies:

• $pc = ab$, so $p = \dfrac{ab}{c}$;
• squaring and using $c^2 = a^2 + b^2$: $\dfrac{1}{p^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2}$.

First find which vertex has the right angle; the side opposite it is the hypotenuse.

Let $ABC$ be right-angled at $A$, with the altitude $AD$ cutting the hypotenuse into $BD = m$ and $DC = n$:

• $AD^2 = BD\cdot DC$, i.e. $p^2 = mn$;
• $AB^2 = BD\cdot BC$ and $AC^2 = DC\cdot BC$ — each leg with the piece next to it and the whole hypotenuse;
• so $BD : DC = AB^2 : AC^2$, and triangles $ABD$ and $ADC$, which share the height $AD$, have areas in that ratio too.

• $h_a = \dfrac{2\Delta}{a}$, and likewise for $b$ and $c$.
• $a : b : c = \dfrac{1}{h_a} : \dfrac{1}{h_b} : \dfrac{1}{h_c}$.
• The smallest altitude stands on the longest side.
• A right triangle inscribed in a circle of radius $R$ has hypotenuse $2R$, so its area is $\dfrac12\times 2R\times p$.