Question

Q. Equilateral triangles are drawn on the sides of a right triangle. Prove that the area of the triangle on the hypotenuse is equal to the sum of the areas of the triangles on the other two sides.

Please solve the above question!! ​

Answer 1

$\huge \underline{ \underline \mathfrak{solution - }}$

Given - A $\triangle \: ABC \:$In which $\angle \: B = 90 \degree$Equilateral triangles$\triangle{BCD \: \triangle CAE \: \triangle \: ABF }$Are drawn on the sides BC, CA and AB respectively.

To prove -

$\{ar( \triangle \: </strong><strong>CAE</strong><strong> </strong><strong>) = ar( \triangle \: </strong><strong>BCD</strong><strong> </strong><strong>) + ar( \triangle \: </strong><strong>ABF</strong><strong> </strong><strong>)}$

Proof -

$\triangle \: </strong><strong>BCD</strong><strong> \: \triangle \: </strong><strong>CAE</strong><strong> \: \triangle \: \:</strong><strong>ABF</strong><strong> </strong><strong>$are equiangular and hence similar.

We have

$\frac{ar( \triangle \: BCD)}{ar( \triangle \: CAE )} + \frac{ar( \triangle \: ABF )}{ ar( \triangle \: cae)} = \frac{ {BC}^{2} }{ {CA}^{2} } + \frac{ {AB }^{2} }{ {CA }^{2} } \\ \implies \frac{ar( \triangle \: BCD ) + ar( \triangle \: ABF )}{ar( \triangle \: CAE )} = \frac{ {BC}^{2} + {AB }^{2} }{ {ca}^{2} } = \frac{ {CA }^{2} }{ {CA }^{2} } = 1$

$\implies \: ar( \triangle \: </strong><strong>BCD</strong><strong>) + ar( \triangle \: </strong><strong>ABF</strong><strong> </strong><strong>) = ar( \triangle \: </strong><strong>CAE</strong><strong> </strong><strong>) \\ \ \underline{ \underline \bold{proved}}$