# Prove that the area of an equilateral triangle described on one side of square is equal to half the area of the equilateral described on one of the diagonals.

## Solution :

Given : A square ABCD, equilateral triangles BCE and ACF have been drawn on side BC and the diagonal AC respectively.

To Prove : $$area(\triangle BCE)$$ = $$1\over 2$$$$area(\triangle DEF)$$

Proof : Since all equilateral triangles are similar.

$$\implies$$  $$\triangle$$ BCE ~ $$\triangle$$ ACF

$$area(\triangle BCE)\over area(\triangle ACF)$$ = $${BC}^2\over {AC}^2$$

Since, Diagonal = $$\sqrt{2}$$ side, So, AC = $$\sqrt{2}$$ BC

$$\implies$$ $$area(\triangle BCE)\over area(\triangle ACF)$$ = $${BC}^2\over {\sqrt{2} BC}^2$$

$$\implies$$  $$area(\triangle BCE)\over area(\triangle ACF)$$ = $$1\over 2$$