Question

In the figure, $AB$ is the diameter of a circle with centre $O$. Chord $DC$ is equal to radius of the circle. $P$ is an external point. Find the measure of $\angle APB$.

Answer 1

Let, $r=$ Radius of Circle.

Here, $\overline{DO}=\overline{CO}=\overline{DC}=r$.

ie, $△OCD$ is an equilateral triangle.

Also, $\overline{DO}=\overline{AO}=r$ and $\overline{CO}=\overline{OB}=r$.

$\therefore$ $△AOD$ & $△BOC$ are isosceles triangles.

Then, $\angle DAO=\angle ADO=\angle A$ and $\angle CBO=\angle BCO=\angle B$.

$\angle DOA={180}^0-(\angle DAO+\angle ADO)$ $(\because$ Angle sum property of triangles.)

$\Rightarrow \angle DOA={180}^0-2\angle A\to$ (1)

Similarly, $\angle COB={180}^0-2\angle B\to$ (2)

$\angle DOC={60}^0$ ($\because$ Angle of an equilateral triangle.)

$\angle AOB={180}^0=\angle DOA+\angle COB+\angle DOC$ ( $\because$ Straight angle)

ie, ${180}^0={180}^0-2\angle A+{180}^0-2\angle B$ (Substituting equations (1) and (2) to above equation)

$\Rightarrow \angle A+\angle B={90}^0\to$ (3)

$\angle APB={180}^0-(\angle DAO+\angle CBO)$

ie, $\angle APB={180}^0-(\angle A+\angle B)={180}^0-{90}^0$ (substituting equation (3))

$\therefore \angle APB={90}^0$