Why is the quadrilateral cyclic

Question

Let ABCD be a cyclic quadrilateral with circumcenter O. Let lines AB and CD meet at E, AD and BC meet at F, and AC and BD meet at P. Furthermore, let EP and AD meet at K, and let M be the projection of O onto AD be M. Prove that BCMK is cyclic.

This is a problem from YufeiZhao.com I was not able to prove, my ideas where the angle KBA=MCD and the intersections being X intersection of BK AC and Y intersection of MC and BD, then XY is paralel to AD but I was not able to prove those supositions.

PLEASE I DON'T WANT A FULL SOLUTION.

Answer 1

We want to show that $$FK\cdot FM = FC\cdot FB= FD\cdot FA.$$ Since $M$ is the midpoint of $AD$, it is equivalent to $$FK=\frac{2FD\cdot FA}{FD+FA},$$ or $$DK=\frac{2FD\cdot FA}{FD+FA}-FD=\frac{FD(FA-FD)}{FD+FA}=\frac{FD\cdot AD}{FD+FA}.$$ Thus we need to show that $$\frac{DK}{AD}=\frac{FD}{FA+FD},$$ equivalently $$\frac{DK}{AK}=\frac{FD}{FA}.$$ Finally, here are the hints:

1. Both $\dfrac{DK}{AK}$ and $\dfrac{FD}{FA}$ are equal to $\dfrac{BD}{CA}\cdot\dfrac{CD}{AB}$.
2. $\dfrac{DK}{AK}$ is given by Menelaus Theorem in $\triangle ACD$ and $\triangle DAB$.
3. $\dfrac{FD}{FA}=\dfrac{FD}{FC}\cdot\dfrac{FC}{FA}$.

Answer 2

Assuming your idea about $\angle KBA = \angle MCD$ is correct,

will $\angle AKB = \angle MCB$?