The given figure shows a pentagon ABCDE in which EG, drawn parallel to DA, meets BA produced at G and CF drawn parallel to DB meets AB produced at F.

Question

The given figure shows a pentagon ABCDE in which EG, drawn parallel to DA, meets BA produced at G and CF drawn parallel to DB meets AB produced at F. Show that ar(pentagon ABCDE) = ar(∆DGF).

Philoid · Accepted Answer

Given: EG||DA, CF||DB

Here, in Quadrilateral ADEG

Area (Δ AED) = Area (Δ ADG) –1

In Quadrilateral CFBD

Area (Δ CBD) = Area (Δ BCF) –2

Add –1 and –2

Area (Δ AED) + Area (Δ CBD) = Area (Δ ADG) + Area (Δ BCF) –3

Add Area (Δ ABD) to –3

Area (Δ AED) + Area (Δ CBD) + Area (Δ ABD) = Area (Δ ADG) + Area (Δ BCF) + Area (Δ ABD)

Area (pentagon ABCDE) = Area (∆DGF)

The given figure shows a pentagon ABCDE in which EG, drawn parallel to DA, meets BA produced at G and CF drawn parallel to DB meets AB produced at F.Show that ar(pentagon ABCDE) = ar(∆DGF).

The given figure shows a pentagon ABCDE in which EG, drawn parallel to DA, meets BA produced at G and CF drawn parallel to DB meets AB produced at F.
Show that ar(pentagon ABCDE) = ar(∆DGF).