Question

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.

Answer 1

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

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

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

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