A circle is tangent to side BC of $\triangle ABC$ at its midpoint $M$, and intersects sides $AB$ and $AC$ at points $E,F$, and $G, H$. Prove that $\displaystyle\frac{BP}{CP} \cdot \frac{BQ}{CQ}=1$.
My attempt:
I used Menelaus' Theorem in $\triangle ABC$ with transversals $P-F-G$ and $Q-H-E$.
$\displaystyle\frac{AF}{BF} \cdot \frac{BP}{CP} \cdot \frac{CG}{AG} = 1$ and $\displaystyle\frac{AE}{BE} \cdot \frac{BQ}{CQ} \cdot \frac{CH}{AH} = 1$.
When these two are multiplied together:
$\displaystyle\frac{AF}{BF} \cdot \frac{BP}{CP} \cdot \frac{CG}{AG} \cdot \frac{AE}{BE} \cdot \frac{BQ}{CQ} \cdot \frac{CH}{AH} =1$.
Then using power of a point:
In the numerator $\rightarrow$ $AF$ and $AE$ cancel out with $AG$ and $AH$, $CG$ and $CH$ cancel out with $BF$ and $BE$ thus leaving us with $\displaystyle\frac{BP}{CP} \cdot \frac{BQ}{CQ} =1$.
Is my attempt correct? Does the fact that $M$ is the midpoint have to be used somewhere?
You can cancel out $CG\cdot CH$ and $BF \cdot BE$ only because: $$CG\cdot CH = CM^2 = BM^2 = BF\cdot BE$$ So the use of $M$ is here (otherwise you can't cancel them out).
Note: The power of point $C$ is given by $CG\cdot CH = CM^2$ and the power of point $B$ is given by $BF\cdot BE=BM^2$, so they are equal if and only if $BM=CM$.