Question

Question 6
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 a quadrilateral having centre A.
To prove that $\angle CBD+\angle CDB=\frac{1}{2}\angle BAD$.
Construction : Join AC and BD.

Proof
The arc DC subtends $\angle DAC$ at the centre and $\angle CBD$ at a point B in the remaining part of the circle.
$\therefore \angle DAC=2\angle CBD$ . . . . . (i)
[In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.]

Similarly, arc BC subtends $\angle CAB$ at the centre and $\angle CDB$ at a point D in the remaining part of the circle.
$\therefore \angle CAB=2\angle CDB$. . . . . (ii)
[In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.]

On adding Eqs. (i) and (ii) 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.