Question

ABC is an isosceles triangle with AB = AC. A line through A meets BC at D and the circumcircle of the triangle at E. Prove that $A{B}^2=AD\cdot AE$.

Answer 1

Given: $AB=AC$

$\angle ABC=\angle ACB–\left(1\right)$

$\angle AEBand\angle ACB$ are made by the same chord $AB$

$\angle ARB=\angle ACB–\left(2\right)$

From $\left(1\right)and\left(2\right)$

$\angle ABC=\angle AEB$

In triangles $ABDandAEB$

$\angle ABC=\angle AEB$

$\angle BAD=\angle BAE$

$⟹ABD$ is similar to $△AEB$

$⟹\frac{AD}{AB}=\frac{AB}{AE}⟹A{B}^2=AD.AE$