ABC is an isosceles triangle where AB = AC which is circumscribed about a circle. If P is the point where the circle touches the side BC, then which of the following is true?
A. PB = PC
B. PB > PC
C. PB < PC
D. PB = $$\frac{1}{2}$$ PC
Answer: Option A
Solution (By Examveda Team)
here given that = AB = AC
AQ + BQ = AR + RC
we know that
BQ = PB & PC = RC
AQ + PB = AR + PC
also AQ = AR
AR + PB = AR + PC
PB = PC
This Question Belongs to Arithmetic Ability >> Triangles
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