Question

Two circles intersect at M and N. The tangent to the first at M meets the second circle at P, while the tangent to the second at N meets the first at Q. Prove that $M{N}^2=NP\cdot QM$

Answer 1

Given: QN is a tangent to 2^nd circle

MP is a tangent to 1^st circle

To show: $\frac{MN}{NP}=\frac{QM}{MN}$

So we have to show $△MNP$ and $△QMN$ are similar.

Since $QN$ and $MP$ are tangents.

According to alternate segment theorem, angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

$\angle NMP=\angle NQM–\left(1\right)$

(MP tangent)

And

$\angle QNM=\angle MPN–\left(2\right)$

(QN tangent)

In triangles $QMN$ and $MNP$

From above (1) and (2) results

$△QMN$ and $△MNP$ are similar (AA similarity)

$⟹\frac{QM}{MN}=\frac{MN}{NP}$

$⟹QM,NP=M{N}^2$