Question

In the figure; PA is a tangent to the circle. PBC is secant and AD bisects angle BAC. Find the value of ∠CAD in terms of ∠PAB and ∠PBA.

Answer 1

Given - In a circle PA is the tangent. PBC is the secant and AD is the bisector of $\angle BAC$ which meets the secant at D.

PA is the tangent and AB is chord.

$\angle PAB=\angle C$

(Angles in the alternate segment)

AD is the bisector is $\angle BAC$

$\therefore \angle 1=\angle 2$

In $\Delta ADC$,

Exterior $\angle ADP=\angle C+\angle 1=\angle PAB+\angle 2=\angle PAD$

$\therefore \Delta PAD$ is an isosceles triangle.

(ii) In $\Delta ABC$,

Exterior $\angle PBA=\angle C+\angle BAC$

$\therefore \angle BAC=\angle PBA-\angle C$

$\Rightarrow \angle 1+\angle 2=\angle PBA-\angle PAB$ [from (i)]

$\Rightarrow 2\angle 1=\angle PBA-\angle PAB$

$\Rightarrow \angle 1=\frac{1}{2}[\angle PBA-\angle PAB]$

$\Rightarrow \angle CAD=\frac{1}{2}[\angle PBA-\angle PAB]$