The given figure shows a pentagon ABCDE. EG drawn parallel to DA meets BA produced at G and CF drawn parallel to DB meets AB produced at F. Prove that the area of pentagon ABCDE is equal to the area of triangle GDF.

We know that,

Triangles on the same base and between the same parallel lines are equal in area.

Since, triangle EDG and EGA lie on the same base EG and between the same parallel lines EG and DA.

∴ Area of △ EDG = Area of △ EGA

Subtracting △ EOG from both sides, we get :

⇒ Area of △ EDG - Area of △ EOG = Area of △ EGA - Area of △ EOG

⇒ Area of △ EOD = Area of △ GOA ………(1)

Since, triangle FDC and FBC lie on the same base FC and between the same parallel lines DB and CF.

∴ Area of △ FDC = Area of △ FBC

Subtracting △ FPC from both sides, we get :

⇒ Area of △ FDC - Area of △ FPC = Area of △ FBC - Area of △ FPC

⇒ Area of △ DPC = Area of △ BPF ………(2)

From figure,

⇒ Area of △ GDF = Area of △ GOA + Area of △ BPF + Area of pentagon ABPDO

⇒ Area of △ GDF = Area of △ EOD + Area of △ DPC + Area of pentagon ABPDO

⇒ Area of △ GDF = Area of pentagon ABCDE.

Hence, proved that area of pentagon ABCDE is equal to the area of triangle GDF.