Question

In the adjoining figure, ABCDE is a pentagon in which EG is drawn parallel to DA meets BA produced at G and CF is drawn parallel to DB meets AB produced at F. Prove that area of ABCDE = area of GDF.

Answer 1

Given: ABCDE is a pentagon with A, B on line GF (GABF collinear). EG is drawn parallel to DA and meets BA produced at G. CF is drawn parallel to DB and meets AB produced at F. D is the top vertex so GDF is the large triangle containing the pentagon as shown.

To Prove: area (ABCDE) = area (GDF).

Proof [Step-wise]:

1. Collinearity and basic decompositions.

Since G, A, B, F are collinear (GABF on one straight line), the big triangle GDF can be split along A and B into three triangles:

area (GDF) = area (ΔDGA) + area (ΔDAB) + area (ΔDBF)

The pentagon ABCDE can be partitioned into three triangles with common vertex D:

area (ABCDE) = area (ΔDEA) + area (ΔDAB) + area (ΔDBC)

So, it is enough to show area (ΔDGA) = area (ΔDEA) and area (ΔDBF) = area (ΔDBC).

2. Equality of the left-hand small triangles.

EG || DA by construction.

Consider the two triangles ΔDGA and ΔDEA.

Using the vector/area or oriented-area expression for triangle area.

area (ΔDGA) = `1/2` |(GD) × (AD)|

And area(ΔDEA) = `1/2` |(ED) × (AD)|

Their difference equals `1/2` |(GE) × (AD)|.

But EG || DA means (GE) is parallel to (AD).

So, the cross product (GE) × (AD) = 0.

Hence, area (ΔDGA) = area (ΔDEA).

3. Equality of the right-hand small triangles.

By the same argument, using CF || DB, we get

area (ΔDBF) = area (ΔDBC)

4. Combine equalities.

Substitute into the decompositions from step 1: 

area (GDF) = area (ΔDGA) + area (ΔDAB) + area (ΔDBF)

= area (ΔDEA) + area (ΔDAB) + area (ΔDBC)

= area (ABCDE)

area (ABCDE) = area (GDF).