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Equal Angles Subtend Equal Sides

In given figu...

Question

In given figure equilateral triangles are drawn on the sides of a right triangle. Show that the area of the triangle on the hypotenuse is equal to the sum of the areas of triangles on the other two sides.

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Solution

Given A right angled triangle ABC with right angle at B. Equilateral triangles PAB, QBC and RAC are described on sides AB, BC and Ca respectively.

To prove $A r e a (△ P A B) + A r e a (△ Q B C) = A r e a (△ R A C)$ .

Proof Since triangles PAB, QBC and RAC are equilateral. Therefore, they are equiangular and hence similar.

$∴$ $\frac{A r e a (△ P A B)}{A r e a (△ R A c)} + \frac{A r e a (△ Q B C)}{A r e a (△ R A C)} = \frac{{A B}^{2}}{{A C}^{2}} + \frac{{B C}^{2}}{{A C}^{2}}$

$\Rightarrow$ $\frac{A r e a (△ P A B)}{A r e a (△ R A C)} + \frac{A r e a (△ Q B C)}{A r e a (△ R A C)} = \frac{{A B}^{2} + {B C}^{2}}{{A C}^{2}}$

$\Rightarrow$ $\frac{A r e a (△ P A B)}{A r e a (△ R A C)} + \frac{A r e a (△ Q B C)}{A r e a (△ R A C)} = \frac{{A C}^{2}}{{A C}^{2}} = 1$

[ $∵ △$ $i s a r i g h t a n g l e d t r i a n g l e w i t h$ $∠ B = 90^{0} ∴ {A C}^{2} = {A B}^{2} + {B C}^{2}$ ]

$\Rightarrow$ $\frac{A r e a (△ P A B) + A r e a (△ Q B C)}{A r e a (△ R A C)} = 1$

$\Rightarrow$ $A r e a (△ P A B) + A r e a (△ Q B C) = A r e a (△ R A C)$ $[H e n c e p r o v e d]$

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