In the parallelogram ABCD, E.F AB.AD And BF = De, BF and de intersect at point P, the bisector angle c of CP is proved

In the parallelogram ABCD, E.F AB.AD And BF = De, BF and de intersect at point P, the bisector angle c of CP is proved

Your question is wrong. It should be CP bisecting angle BPD. It is impossible to bisecting angle C. then it is right and equal according to the solution of his question. Connect EC and FC, the area of triangle ECD and triangle BCF is equal, which is equal to half of the area of parallelogram