$P A$ and $P B$ are tangents from $P$ to the circle with center $O$ . At point $M$ , a tangent is drown cutting $P A$ at $K$ and $P B$ at $N$ . Prove that $K N = A K + B N$ .

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Solution

We know that the tangents drawn from an external point to a circle are equal in length.

$∴ P A = P B [F r o m P] (i)$

$K A = K M [F r o m K] (i i)$

and, $N B = N M [F r o m N] (i i i)$

Adding $(i i), (i i i)$ ,we get

$K A + N B = K M + N M$

$\Rightarrow A K + B N = K M + M N$

$\Rightarrow A K + B N = K N$

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PA and PB are tangents from P to the circle with center O. At point M, a tangent is drown cutting PA at K and PB at N. Prove that KN=AK+BN.

We know that the tangents drawn from an external point to a circle are equal in length.∴PA=PB [From P](i)KA=KM[From K](ii)and, NB=NM[From N](iii)Adding (ii),(iii),we getKA+NB=KM+NM⇒AK+BN=KM+MN⇒AK+BN=KN

$P A$ and $P B$ are tangents from $P$ to the circle with center $O$ . At point $M$ , a tangent is drown cutting $P A$ at $K$ and $P B$ at $N$ . Prove that $K N = A K + B N$ .