Question

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

Answer 1

Let square $ABCD$ and triangle $ABE$ is described on side $AB$ and triangle $ACF$ is described on diagonal $AC$ you can take other sides also.

Now triangle $ABE$ is similar to triangle $ACF$ because both are equilateral triangle

By area theorem

$\frac{Areaof△ABE}{Areaof△ACF}={\left(\frac{AB}{AC}\right)}^2$......(1)

By Pythagoras theorem in $△ABC$

$A{B}^2+B{C}^2=A{C}^2$

Sides of a square are equal,

$\therefore 2A{B}^2=A{C}^2$

${\left(\frac{AB}{AC}\right)}^2=\frac{1}{2}$

now putting this value in (1)

$\frac{Areaof△ABE}{Areaof△ACF}=\frac{1}{2}$

$2\times Areaof△ABE=Areaof△ACF$