Question

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.Show that ar(pentagon ABCDE)=ar($△DGF$).

Answer 1

Given: ABCDE is a pentagon. EG || DA and CF || DB.
To prove: ar(pentagon ABCDE ) = ar ( ∆ DGF)
Proof:
ar(pentagon ABCDE )​ = ar(∆ DBC) + ar(∆ ADE ) + ar(∆ ABD)
Also, ar (∆ DGF) = ar(∆ DBF) + ar(∆ ADG) + ar(∆ ABD )
...(i)
...(ii)
Now, ∆DBC and ∆ DBF lie on the same base and between the same parallel lines.
∴ ar(∆ DBC) = ar(∆ DBF)
...(iii)
Similarly, ∆ADE and ∆ADG lie on same base and between the same parallel lines.
∴ ar(∆ ADE) = ar(∆ ADG)
...(iv)
From (iii) and (iv), we have:
ar(∆ DBC) + ar(∆ ADE) = ar(∆ DBF) + ar(∆ ADG)
Adding ar(∆ ABD) on both sides, we get:
ar(∆ DBC) + ar(∆ ADE) + ar(∆ ABD) = ar (∆ DBF) + ar(∆ ADG) + ar(∆ ABD)
By substituting the values from (i) and (ii), we get:
ar(pentagon ABCDE) = ar (∆ DGF)