Passing through point E and point G respectively is the height of the common edge of △PFE and △PFG on FP. It is proved that the right triangle GKP and BPE are similar, and AGP and PKE are similar. It is proved that GK=HE is achieved through the transformation of AP=PB, and the areas of triangles with equal bottoms and equal heights are equal.

Prove:

EH⊥PF and GK⊥PF pass through point E and point G respectively, and the vertical foot is point H and point K respectively.

PF⊥DC、pe⊥bc==>； ∠FPE+∠C= 180 degrees

In the circle inscribed quadrilateral ABCD, ∠A+∠C= 180 degrees.

So ∠FPE=∠C

pg⊥ad,eh⊥pf==>； ∠PGA=∠EHP=90 degrees

So, △AGP∽△PHE

Therefore, GP/HE=AP/PE.

It can also be proved that: △GKP∽△PEB.

Therefore, BP/PE=GP/GK.

P is the midpoint of AB == >; AP=PB

So GP/ he =GP/GK

So he =GK

Because S△PFG= 1/2*PF*GK.

S△PFE= 1/2*PF*HE

So S△PFG=S△PFE