Question

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

Answer 1

Consider the quadrilateral ADEG.

Let the intersection of AE and DB be K.

So triangles ADE and ADG have the same base AD and the same height, since parallel lines are equal, distances apart, therefore, $△ADE=△ADG$.

But triangle ADK is common to both, therefore, $△AKG=△DEK$.

Similarly by letting FD and BC intersect at L, we can show that $△FLB=△CDL$.

Now, pentagon ABCDE = $△BLD+△LCD+△ABD+△DEK+△ADK$

=$△BLD+△FLB+△ABD+△GAK+△ADK$

=$△GDF$

Hence proved.