Question

In the given figure ABC is a triangle in which AB=AC. Points D and E are points on the sides AB and AC respectively such that AD=AE. Show that the points B,C,E and D are concyclic.

Answer 1

In order to prove that the points B,C,E and D are concyclic, it is sufficient to show that $\angle ABC+\angle CED=180$0$and$\angle ACB+\angle BDE=180^0$.

In $△ABC$, we have

$AB=AC$ and $AD=AE$

$\Rightarrow$ $AB-AD=AC-AE$

$\Rightarrow$ $DB=EC$

Thus, we have

$AD=AE$ and $DB=EC$

$\Rightarrow$ $\frac{AD}{DB}=\frac{AE}{EC}$

$\Rightarrow$ $DE||BC$ [By the converse of Thale's Theorem]

$\Rightarrow$ $\angle ABC=\angle ADE$ [Corresponding angles]

$\Rightarrow$ $\angle ABC+\angle BED=\angle ADE+\angle BDE$ [Adding $\angle BDE$ both sides]

$\Rightarrow$ $\angle ABC+\angle BDE={180}^0$

$\Rightarrow$ $\angle ACB+\angle BDE={180}^0$ [$\because AB=AC\therefore \angle ABC=\angle ACB$]

Again, $DE||BC$

$\Rightarrow$ $\angle ACB=\angle AED$

$\Rightarrow$ $\angle ACB+\angle CED=\angle AED+\angle CED$ [Adding $\angle CE$ on both sides]

$\Rightarrow$ $\angle ACB+\angle CED={180}^0$

$\Rightarrow$ $\angle ABC+\angle CED={180}^0$ [$\because \angle ABC=\angle ACB$]

Thus, BDEC is a cyclic quadrilateral. Hence, B,C,E and D are concyclic points.