Similar triangles and circle

Question

We have $k(O;r$). $AB$ is a chord and $t_1$ and $t_2$ are tangents to the circle at points $A$ and $B$, respectively. I should show that the distance from every point lying on the circle to $AB$ is equal to the product of the distances from the point to $t_1$ and $t_2$.

So, if we succeed in showing $\dfrac{PQ}{PM}=\dfrac{PN}{PQ}$, the problem is solved. We can try to show $\triangle MPQ \sim \triangle QPN$, but I don't see how to. I am trying to use the fact that $AQPM, QBNP$ and $MPNK$ are cyclic quadrilaterals, but it seems useless at the end. Can you help me?

Answer 1

Here's a hint:

$$\angle PMQ = \angle PAQ = \angle PAB = \angle PBN = \angle PQN.$$

Answer 2

Note that $\angle KAB = \angle KBA$ due to the tangent points $A$ and $B$. Then, the two cyclic quadrilaterals $AMPQ$ and $BNPQ$ are similar. Thus,

$$\dfrac{PQ}{PM}=\dfrac{PN}{PQ}$$