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