prove that the area of an …

Given:  A square ABCD an equilateral triangle ABC and ACF have been described on side BC and diagonal AC respectively.

To Prove :-  {tex}ar( ∆BCE ) = \bf{ \frac{1}{2} ar( \triangle ACF ) . }{/tex}

Proof :-  Since each of the ∆ABC and ∆ACF is an equilateral triangle, so each angle of his strength is one of them is 60°. So, the angles are equiangular, and hence similar.

 => ∆BCE ~ ∆ACF.

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}\bf{ \frac{ ar( \triangle BCE ) }{ ar( \triangle ACF ) } = \frac{{BC}^{2} }{ {AC}^{2}} = \frac{{BC}^{2}}{2{(BC)}^{2}}. }{/tex}

[ Because, AC is hypotenuse => AC = √2BC. ]

{tex}\bf{ \implies \frac{ ar( \triangle BCE ) }{ ar( \triangle ACF ) } = \frac{1}{2} . }{/tex}

Hence,  {tex}\boxed{ \sf ar( \triangle BCE ) = \frac{1}{2} \times ar( \triangle ACF ) . }{/tex}