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.triangles

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

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

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

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