CP=2AP

[ins-]

Triangle ABC is inscribed in a circle k. The height from C intersects AB and k at the points D and E respectively. F is the middle of the segment BD. A line l through D, perpendicular to EF intersects the segment AC at the point P. Prove that CP=2AP.

[Ilya Nek]

Let $G$ is the middle of segment $FC$. Then $l$ is the median of triangle $AFG$(because 1.$EFGA$ is inscribed in a circle 2. orthocenter is isogonal conjugates to circumcenter 3. Circumcenter in rectangular triangle = the middle of hypotenuse ). Menelaus for triangle $AGC$ and line $DP$ gives $CP=2AP$.

[Liouville]

Пусть $K$ - точка симметричная точке $D$ относительно $A$. Так как $FD\cdot{DK}=\frac{BD}{2}\cdot{2DA}=ED\cdot{DC}$, то $FCKE$ - вписанный. Пусть $DP$ пересекает $FE$ и $CK$ в точках $M$ и $N$, соответственно. Тогда $\angle{NCD}=\angle{MFD}=\angle{MDE}=\angle{NDC}$, следовательно $N$ - середина $CK$, а медиана $DN$ треугольника $CDK$ делит другую его медиану $CA$ в отношении 2:1.

[ins-]

Thank you guys. You the problem is easy but I hope nice. You can see more approaches to solve it here:
http://www.artofproblemsolving.com/Foru ... 9#p2458109
http://www.math10.com/f/viewtopic.php?f ... 759#p35759