Call the point of concurrency O. (1) We willl show that: triangle AEC ~ triangle CDO i.angle AEC = angle ADC because of congruent intercepting arcs ii.angle EAC= angle AED+angle DAC= angle ECD+angle ECF=angle DCO Therefore, by AA, triangle AEC ~ triangle CDO AC/CE=CO/DO (2) We will show that: triangle AEC ~ triangle CDO i.angle ACE = angle ADE because of congruent intercepting arcs ii.angle AEO =angle CED because AB=CD Adding BEC to both sides yields: angle AEC= angle OED AC/CE=OD/DE Multiplying yields: (AC/CE)^2=CO/DE Triangle DXE ~ triangle OXC is easy to show. Therefore OC/CE=CX/EX=(AC?CE)^2

Call the point of concurrency O. (1) We willl show that: triangle AEC ~ triangle CDO i.angle AEC = angle ADC because of congruent intercepting arcs ii.angle EAC= angle AED+angle DAC= angle ECD+angle ECF=angle DCO Therefore, by AA, triangle AEC ~ triangle CDO AC/CE=CO/DO (2) We will show that: triangle AEC ~ triangle CDO i.angle ACE = angle ADE because of congruent intercepting arcs ii.angle AEO =angle CED because AB=CD Adding BEC to both sides yields: angle AEC= angle OED AC/CE=OD/DE Multiplying yields: (AC/CE)^2=CO/DE Triangle DXE ~ triangle OXC is easy to show. Therefore OC/CE=CX/EX=(AC?CE)^2

Just for the sake of completeness: the problem was discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=21738 . darij

Let $AD,BE,CF$ meet at $X.$ Angle chasing gives $\triangle ACE\sim\triangle XBA\sim\triangle FEX.$ We also have $\triangle EXP\sim\triangle EBC.$ Thus \[\left(\frac{AC}{CE}\right)^2=\left(\frac{XB}{BA}\right)\left(\frac{FE}{EX}\right)=\frac{XB}{EX}=\frac{CP}{PE}.\]