if your diagram is right, then BC = BP is wrong.
THe diagram shows triangle ABC where points A , B , C and P are coplanar . BC=BP=AP and angle BAC =angle ABP = angle PBC = pi/5 rad
Prove that triangle ABC and BPC are similar triangles . Hence deduce that
(1) BC^2=CP.CA
(2)
I am only stucked with (2) , i am ok with the rest . THanks .