Question

In the given figure, AD = AE and $A{D}^2=BD\times EC$.

Hence , $△ABD$ $\sim$ $△CAE$.

If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

GIven, $AD=AE$

$\angle ADE=\angle AED$ (Isosceles triangle propery)

$180-\angle ADE=180-\angle AED$

$\angle ADB=\angle AEC.....\left(1\right)$

Given: $A{D}^2=BD\times EC$

$\frac{AD}{EC}=\frac{BD}{AD}...\left(2\right)$

Now, in $△ABD$ and $△ACE$,

$\angle ADB=\angle AEC$ (From $\left(1\right)$)

$\frac{AD}{EC}=\frac{BD}{AD}$ (From $\left(2\right)$)
$\frac{AD}{BD}=\frac{EC}{AE}$

Corresponding sides of two triangles are proportional.

Thus, $△ABD\sim △ACE$ (SAS criterion)