Similar Figures on Sides of Right-Angled Triangle

Theorem

As Euclid defined it:

In right-angled triangles the figure on the side subtending the right angle is equal to the similar and similarly described figures on the sides containing the right angle.

(The Elements: Book VI: Proposition $31$)

Proof

Let $\triangle ABC$ be a right-angled triangle where $\angle BAC$ is a right angle.

We need to show that the area of the figure on $BC$ is equal to the sum of the areas of the similar figures on $AB$ and $AC$.

Let $AD$ be drawn perpendicular to $BC$.

From Perpendicular in Right-Angled Triangle makes two Similar Triangles, $\triangle ABD$ and $\triangle ADC$ are similar both to each other and to $\triangle ABC$.

So from Book VI Definition 1: Similar Rectilineal Figures, $CB : BA = AB : BD$.

So from the porism to Ratio of Areas of Similar Triangles, the figures on each of these sides are proportional in the same way.

So $BC : BD$ is the same as the ratio of the areas of the figures on $CB$ and $BA$.

Similarly, $BC : CD$ is the same as the ratio of the areas of the figures on $CB$ and $CA$.

So $BC : BD + DC$ equals the ratio of the areas of $BC$ and the sum of the areas of the figures on $CA$ and $AB$.

$\blacksquare$

Historical Note

This is Proposition 31 of Book VI of Euclid's The Elements.

Also see Book I Proposition 47: Pythagoras's Theorem, of which this is an extension.