Generalization of Pythagoras’ theorem

In the following figure the triagle $\mathrm{AB} \Gamma$ is right angled. Using the sides of the triangle we draw equilateral triangles as shown in the picture.

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Prove that

$\mathrm{E}_1 = \mathrm{E}_2 + \mathrm{E}_3$

Solution

Since $\mathrm{AB} \Gamma$ is right angled , we know that Pythagoras’ theorem applies; hence:

$\mathrm{AB}^2 + \mathrm{A} \Gamma^2 = \mathrm{B}\Gamma^2$

On the other hand the area of an equilateral triangle of side $\alpha$ is given by the formula $\displaystyle \mathrm{E} = \frac{\alpha^2 \sqrt{3}}{4}$ . Hence,

$\mathrm{AB}^2 + \mathrm{A} \Gamma^2 = \mathrm{B}\Gamma^2 \Leftrightarrow \frac{\sqrt{3}\mathrm{AB}^2}{4} + \frac{\sqrt{3} \mathrm{A} \Gamma^2}{4} = \frac{\sqrt{3} \mathrm{B}\Gamma^2}{4} \Leftrightarrow \mathrm{E}_3 + \mathrm{E}_2 = \mathrm{E}_1$