Prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal.

Given: An equilateral triangle described on one side of the square.

To do: To prove that the area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal.

 Let a be the side of square.

 side of $\vartriangle ABC =a$

Area of the $\vartriangle ABC=\frac{1}{2}\times\ side\times\ side\times\  sin\theta$

                 $=\frac{1}{2}×a×a×sin60^{o}$                $( \because \vartriangle ABC\ is\ an\ equilateral\ \therefore \theta=60^{o})$

                                 $=\frac{\sqrt{3}}{4}a^{2}$

$\vartriangle AED$, is described on the squre 's diagonal.

diagonal $=\sqrt{a^{2}+a^{2}}=a\sqrt{2}$                 $ ( using\ pythagoras\ theorem)$

Area of $\vartriangle AED=\frac{\sqrt{3}}{4}(a\sqrt{2})^{2}$

                                   $=\frac{\sqrt{3}}{4}(2a^{2})$

$\therefore \frac{ar( \vartriangle ABC)}{ar( \vartriangle AED)}=\frac{\frac{\sqrt{3}}{4}a^{2}}{\frac{\sqrt{3}}{4}(2a^{2})}=\frac{1}{2}$

Hence it has been proved that  area of an equilateral triangle described on one side of the square is equal to half the area of the equilateral triangle described on one of its diagonal.