CGMO 2005 Problem 1

As shown in the following figure, point $P$ lies on the circumcicle of triangle $ABC.$ Lines $AB$ and $CP$ meet at $E,$ and lines $AC$ and $BP$ meet at $F.$ The perpendicular bisector of line segment $AB$ meets line segment $AC$ at $K,$ and the perpendicular bisector of line segment $AC$ meets line segment $AB$ at $J.$ Prove that

$$\left(\frac{CE}{BF} \right)^2 = \frac{AJ \cdot JE}{AK \cdot KF}.$$

Attachments: