Question

In the figure, $PA$ is a tangent to the circle. $PBC$ is secant and $AD$ bisects angle $BAC$. Select the statements that are true.

• A. $\angle CAD=\frac{1}{3}(\angle PBA-\angle PAB)$
• B. Triangle PAD is an isoscles triangle
• C. $\angle CAD=\frac{1}{2}(\angle PBA-\angle PAB)$
• D. $\angle CAD=\frac{1}{9}(\angle PBA-\angle PAB)$

Answer 1

Answer: (B), (C)

The correct options are

B Triangle PAD is an isosceles triangle

C $\angle CAD=\frac{1}{2}(\angle PBA-\angle PAB)$

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.

$\angle PAB=\angle ACP$ (Angles in the alt. segment)...(i)

AD is the bisector is $\angle BAC$

$\therefore \angle 1=\angle 2$...(ii)

In $\Delta ADC$,

$\angle ADP=\angle ACP+\angle 1=\angle PAB+\angle 2=\angle PAD$ (From (i)and (ii)]

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

(ii) In $\Delta ABC$,

$\angle PBA=\angle ACP+\angle BAC$

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

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

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

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

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