Similar Figures on Sides of Right-Angled Triangle

Theorem

 * 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.

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 $AC$.


 * Euclid-VI-31.png

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, $CB : BA = AB : BD$.

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 $AB$.