Prove that the area of the equilateral triangle drawn on the hypotenuse of a right-angled triangle is equal to the sum of the areas of the equilateral triangles drawn on the other two sides of the triangle.

#### Solution

Let a right triangle BAC in which ∠A is right angle and AC = y, AB = x.

Three equilateral triangles ΔAEC, ΔAFB and ΔCBD are drawn on the three sides of ΔABC.

Again let area of triangles made on AC, AS and BC are A_{1}, A_{2} and A_{3}, respectively.**To prove:** A_{3} = A_{1} + A_{2}

**Proof:** In ΔCAB, by Pythagoras theoren,

`BC^2 = AC^2 + AB^2`

⇒ `BC^2 = y^2 + x^2`

⇒ `BC = sqrt(y^2 + x^2)`

We know that, area of an equilateral triangle = `sqrt(3)/2` (Side)^{2}

∴ Area of equilateral ΔAEC, `A_1 = sqrt(3)/4 (AC)^2`

⇒ `A_1 = sqrt(3)/4 y^2` .....(i)

And area of equilateral ΔAFB, `A_2 = sqrt(3)/2 (AB)^2`

= `sqrt(3)/4 sqrt((y^2 + x^2))`

= `sqrt(3)/4 (y^2 + x^2)`

= `sqrt(3)/4 y^2 + sqrt(3)/4 x^2`

= A_{1} + A_{2} ......[From equations (i) and (ii)]

Hence proved.