Similar Figures on Sides of Right-Angled Triangle

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Theorem

In the words of Euclid:

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 $\text{VI}$: 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 $AC$.

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 Book $\text{VI}$ 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 proof is Proposition $31$ of Book $\text{VI}$ of Euclid's The Elements.
Also see Proposition $47$ of Book $\text{I} $: Pythagoras's Theorem, of which this is an extension.


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