Question

D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE.

Prove that the points B, C, E and D are concyclic.

Answer 1

In $△ABC$, $AB=AC\dots \dots \left(i\right)$

Also given $AD=AE$

Therefore $\angle ABC=\angle ACB=x$ (say)

In $△ADE,AD=AE\dots \dots \left(ii\right)$

Now in $△ABC$, we have

$\frac{AD}{AB}=\frac{AE}{AC}$ [Dividing eq.(ii) by eq.(i)]

$\Rightarrow DE\parallel BC$ [by Converse of BPT ]

$\Rightarrow \angle BCE=\angle DEA=\angle ADE=x$

Therefore $\angle BCE+\angle BDE=x+({180}^{\circ }-x)={180}^{\circ }$ [$\because \angle BDE={180}^{\circ }-\angle ADE={180}^{\circ }-x$]

So, $BCED$ is a quadrilateral in which sum of opposite angle is ${180}^{\circ }$

$\Rightarrow BCED$ is a cyclic quadrilateral and the points $B,C,E$ and $D$ are concyclic.