Question

in fig , the sides AB,BC and CA of triangle ABC touch a circle with centre O and radius r at points P , Q, and R respectively. prove that :

1) AB + CQ = AC + BQ
2) ar(∆ ABC) = 1/2(perimeter of ∆ ABC) * r

Answer 1

Answer is angle A and in it Angle p bisect q

Answer 2

Let

A= Area of triangle ABC

A= Area of triangle BOC + Area of triangle AOC + Area of triangle AOB

Here sides of triangle are base and r radius becomes height to each sub triangle

A=1/2(AB*r)+1/2(BC*r)+1/2(CA*r)

A=1/2(AB+BC+CA)r

A=1/2(perimeter of ABC)*r

Using the property, lengths of tangent from an external point are equal in lengths,

AP = AR   → (1)

BP = BQ   → (2)

CQ = CR  →  (3)

Adding (1) and (2),

AB = AR + BQ  → (4)

Adding (1) and (3),

AP + CQ = AC    

⇒AC = AP + CQ   → (5)  

Subtracting (4) from (5),

AC - AB = AP + CQ - AR - BQ

Since AP = AR,

AC -AB = CQ -BQ

AC + BQ = AB + CQ