ABC is an isosceles triangle right angled at B two equilateral triangles are constructed

>

ABC is an isosceles triangle right angled at B. Similar triangles ACD and ABE are constructed on sides AC and AB. Find the ratio between the areas of ABE and ACD .

Solution

Let AB = BC = x

It is given that $△$ABC is right-angled at B

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

$\Rightarrow$ $A{C}^2=x^2+x^2$

$\Rightarrow$ AC = $\sqrt{2}$x

Its given that

$△$ABE ~ $△$ACD

$\Rightarrow$ $\frac{Area\left(△ABE\right)}{Area\left(△ACD\right)}=\frac{A{B}^2}{A{C}^2}$

$\Rightarrow$ $\frac{Area\left(△ABE\right)}{Area\left(△ACD\right)}=\frac{x^2}{(\sqrt{2}x{)}^2}$

$\Rightarrow$ $\frac{Area\left(△ABE\right)}{Area\left(△ACD\right)}=\frac{1}{2}$