Question

Let $ABC$ be a triangle with circumcentre $O$. The points $P$ and $Q$ are interior points of the sides $CA$ and $AB$, respectively. Let $K,L$ and $M$ be the mid points of the segments $BP,CQ$ and $PQ$ respectively, and let $\tau$ be the circle passing through $K,L$ and $M$. Suppose that the line $PQ$ is tangent to the circle $\tau$. Prove that $OP=OQ$.

Answer 1

From $AB\parallel KM$ and $AC\parallel LM$, we obtain that

$\angle KMQ\equiv \angle AQP.....\left(i\right)$

$\angle LMP\equiv \angle APQ$

If PQ is a tangent to the circumcircle of $△KLM$, then

$\angle KLM\equiv \angle KMQ$

$\therefore \angle KMQ\equiv \angle AQP\left[\text{From}\left(i\right)\right]$

Then $△APQ\sim △MKL$

$\therefore \frac{AQ}{ML}=\frac{AP}{MK}$

It follows that

$\frac{AQ}{PC}=\frac{AP}{BQ}$

$\Rightarrow AQ.BQ=AP.PC$

This means that P and Q have equal powers w.r.t. the circumcircle of the $△ABC$.

As they are both situated inside the circle, they are at equal distance from the centre of the circle O.

Hence, $OP=OQ$