Prove that area of an equilateral …

Given: A square ABCD and equilateral triangle BCE and ACF have been described on the side BC and diagonal AC respectively.
To Prove {tex}\operatorname { ar } ( \Delta B C E ) = \frac { 1 } { 2 } \operatorname { ar } ( \Delta A C F ){/tex}.
Proof: Since each of the {tex}\triangle BCE{/tex} and {tex}\triangle ACF{/tex} is an equilateral triangle, so each angle of each one of them is 60°.
So, the triangles are equiangular and hence similar.
{tex}\therefore \quad \triangle B C E \sim \triangle A C F{/tex}
We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.
{tex}\therefore \quad \frac { \operatorname { ar } ( \triangle B C E ) } { \operatorname { ar } ( \triangle A C F ) } = \frac { B C ^ { 2 } } { A C ^ { 2 } } = \frac { B C ^ { 2 } } { 2 ( B C ) ^ { 2 } }{/tex} {tex}[ \because A C = \sqrt { 2 } B C ]{/tex}
{tex}= \frac { 1 } { 2 }{/tex}
Hence, {tex}\operatorname { ar } ( \triangle B C E ) = \frac { 1 } { 2 } \times \operatorname { ar } ( \triangle A C F ){/tex} .