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

we know that angle subtended by the arc at the center is twice the angle subtended by the arc on the circle

therefore $\angle CDB=\frac{\angle BAC}{2}$ ......eq(1)

$\angle CBD=\frac{\angle DAC}{2}$ .........eq(2)

Adding equation (1) and (2)

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