Question

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides. Prove that the quadrilateral, so formed is cyclic.

Answer 1

Given $\Delta ABC$ is an isosceles triangle such that AB = AC and also DE || BC.
To prove that quadrilateral BCDE is a cyclic quadrilateral.
Construction : Draw a circle passes through the points B, C, D and E

Proof
In $\Delta ABC$,
AB = AC [equal sides of a isosceles triangle]
$\angle ACB=\angle ABC$ . . . . . .(i) [angles opposite to the equal sides are equal]
Since,DE || BC
$\angle ADE=\angle ACB$ . . . . . .(ii)
On adding $\angle EDC$ on both sides in Eq. (ii), we get
$\angle ADE+\angle EDC=\angle ACB+\angle EDC$
$\Rightarrow {180}^{\circ }=\angle ACB+\angle EDC$
$\Rightarrow \angle EDC+\angle ABC={180}^{\circ }[fromEq.(i\left)\right]$
Hence, BCDE is a cyclic quadrilateral, because sum of the opposite angles is ${180}^{\circ }$.