PA and PB are tangents from P to the circle with center O. At point M which lies on circle, a tangent is drawn cutting PA at K and PB at N. Prove that KN = AK + BN. [2 MARKS]
Concept: 1 Mark
Application: 1 Mark
We know that the tangents drawn from an external point to a circle are equal in length.
KA = KM .... (i) [From K]
and, NB = NM .... (ii) [From N]
Adding equations (i) and (ii), we get
KA + NB = KM + NM
⇒AK+BN=KM+MN
⇒AK+BN=KN