In Fig. 10.83, if AP = PB, then

Given:

AP = PB

Property: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By the above property,

AP = AQ (tangent from A)

BR = BP (tangent from B)

CQ = CR (tangent from C)

Clearly,

AP = BP = BR

AQ = AP = BR

Now,

AQ + QC = BR + RC

⇒ AC = BC [∵AC = AQ + QC and BC = BR + RC]

Hence, AC = BC