Question

In the figure, A is the centre of the circle. ABCD is a parallelogram and CDE is a straight line. Find $\angle BCD:\angle ABE$.

Answer 1

In the figure, ABCD is a parallelogram and CDE is a straight line.

$\because$ ABCD is a ||gm

$\therefore \angle A=\angle C$

and $\angle C=\angle ADE$ (Corresponding angles)

$\Rightarrow \angle BCD=\angle ADE$

Also, $\angle ABE=BED$ (Alternate angles)

$\because$ arc BD subtends $\angle BAD$ at the centre and $BED$ at the remaining part of the circle.

$\therefore \angle BED=\frac{1}{2}\angle A=\frac{1}{2}\angle C=\frac{1}{2}\angle ADE$

Now, $\frac{\angle BCD}{\angle ABE}=\frac{\angle BAD}{\angle BED}=\frac{\angle ADE}{\frac{1}{2}\angle ADE}=\frac{2}{1}$

$\therefore \angle BCD:\angle ABE=2:1$