Question

$ABCD$ is such a quadrilateral 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

Given: In a circle, $ABCD$ is quadrilateral having centre $A$

To prove: $\angle CBD+\angle CDB=\frac{1}{2}\angle BAD$

Construction: Join $AC$ and $BD$

Proof: Since, arc $DC$ subtends $\angle DAC$ at the centre and $\angle CBD$ at a point $B$ in the remaining part of the circle

$\therefore$ $\angle DAC=2\angle CBD.....\left(1\right)$ (Angle subtended by an arc at the centre is twice the angle subtended by it on circumference)

Similarly, arc $BC$ subtends $\angle CAB$ at the centre and $\angle CDB$ at a point $D$ in the remaining part of the circle

$\therefore$ $\angle CAB=2\angle CDB.......\left(2\right)$ (Angle subtended by an arc at the centre is twice the angle subtended by it on circumference)

On adding Eqs. $\left(1\right)$ and $\left(2\right)$ we get

$\angle DAC+\angle CAB=2\angle CBD+2\angle CDB$

$\Rightarrow$ $\angle BAD=2(\angle CBD+\angle CDB)$

$\Rightarrow$ $\angle CDB+\angle CBD=\frac{1}{2}\angle BAD$

Hence proved.