Question

ABCD is a quadrliateral such that A is the centre of the circle passing through B, C and D. Prove that $\angle CBD+\angle CDB=\frac{1}{2}\angle BAD$

Answer 1

$A$ is the center of a circle passing through the points $B,C,$ and $D$ of a quadrilateral.

Join $AE$ and $BE.$

We know that the angle subtended by a chord at the center of a circle is double the angle subtended by it at the remaining part of the circle.

So, $\angle BAD=2\angle BED$

Also, $EBCD$ is a cyclic quadrilateral,

$\angle BED+\angle BCD={180}^{\circ }$ (opposite angles of a cyclic quadrilateral are supplementary)

$\Rightarrow \angle BED={180}^{\circ }–\angle BCD$

$\Rightarrow \angle BAD={360}^{\circ }–2\angle BCD$

$\Rightarrow \angle BCD=\frac{{360}^{\circ }-\angle BAD}{2}={180}^{\circ }-\frac{\angle BAD}{2}$

In $△BCD,$ by using angle sum property, we have

$\angle CBD+\angle CDB+\angle BCD={180}^{\circ }$

$\Rightarrow \angle CBD+\angle CDB={180}^{\circ }–\angle BCD$

$={180}^{\circ }-[{180}^{\circ }-\frac{\angle BAD}{2}]$

$=\frac{1}{2}\angle BAD$

Hence proved.