In the figure below, the incircle of ΔPQR touches the sides QR, RP and PQ at K, L and M respectively. If PQ = PR, prove that KQ = KR

Theorem: The lengths of the tangents drawn from an external point to a circle are equal.

Therefore, from the above figure we have,

PM = PL

MQ = KQ

KR = LR

Adding them we get,

PM + MQ + KR = PL + KQ + LR

PQ + KR = PR + KQ

[Since, PM + MQ = PQ and PL + LR = PR]

[Also, PQ = PR(Given)]

KR = KQ

Hence, Proved.