Question

PA is tangent to the circle with center o. PBC is secant and AD bisects angle BAC. Show that triangle PAD is isoceles triangleand angle CAD=1/2*(anglePBA-anglePAB).

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 triangle ADC 

Ext.  

Angle ADP = Angle C + Angle 1  

= Angle PAB + Angle 2 = Angle PAD 

Triangle PAD is an isosceles triangle  

2) In triangle ABC  

Ext. 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]

Answer 2

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 triangle ADC 

Ext.  

Angle ADP = Angle C + Angle 1  

= Angle PAB + Angle 2 = Angle PAD 

Triangle PAD is an isosceles triangle  

2) In triangle ABC  

Ext. 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]