Question

In teh figure; PA is a tangent to the circle, PBC is secant and AD bisects angle BAC. Show that triangle PAD is an isosceles triangle. Also, show that :

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

Answer 1

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


1) PA is the tangent and AB is chord


$\angle PAB=\angle C$ (angles are alternate segment)


AD is the bisector of angle BAC


2) $\angle 1=\angle 2$


$In△ADC,$

$\angle ADP=\angle C+\angle 1$


$=\angle PAB+\angle 2=\angle PAD$


Triangle PAD is an isosceles triangle


2) In triangle ABC


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


$\angle BAC=\angle PBA–\angle C$


$\angle 1+\angle 2=\angle PBA–\angle PAB$


From equation 1


2 $\angle 1=\angle PBA–\angle PAB$


$=\angle 1=½[\angle PBA-\angle PAB]$


$\angle CAD=½[\angle PBA-\angle PAB]$