PA and PB are tangents from P to the circle with center O. At point M, a tangent in drawn cutting PA at K and PB at N. Prove that KN = AK + BN.
Dear student,
following is the answer to your question:
In the given figure,
PA = PB .......(length of tangent from an external point to a circle are equal)
similarly, KA = KM and ..............(1)
NB = NM .............(2)
adding (1) and (2) we get,
KA + NB = KM + NM
=> AK + NB = KN (since, KM + NM = KN)
Hence proved.