Question

In the adjoining figure, AD = BD and ◻BDEC is a parallelogram. Prove that area (ΔABC) = area (◻BDEC).

Answer 1

Given: AD = BD, so D is the midpoint of AB and ◻BDEC is a parallelogram.

To Prove: area (△ABC) = area (◻BDEC).

Proof [Step-wise]:

1. From AD = BD, D is the midpoint of AB.

2. Let h(X) denote the perpendicular distance from point X to the line BC.

Then h(B) = 0 (since B lies on BC) and write h(A) for the altitude of A to BC.

3. Points A, D, B are collinear with D the midpoint.

Distances to a fixed line vary linearly along a segment, so the distance of the midpoint to the line is the average of the endpoint distances.

Hence, `h(D) = (h(A) + h(B))/2`

= `(h(A) + 0)/2`

= `(h(A))/2`

One can justify this by similar triangles obtained from drawing parallels to BC through A and B.

4. `"Area of" △ABC = 1/2 xx (BC) xx h(A)`.

Area of parallelogram BDEC

= (base BC) × (height from D to BC) 

= (BC) × h(D)

Recall: area of a parallelogram = base × height.

5. Substitute `h(D) = (h(A))/2` into the parallelogram area:

Area (◻BDEC) = BC × h(D)

= `BC xx (h(A))/2` 

= `1/2 xx BC xx h(A)`

= Area (△ABC)

Area (△ABC) = Area (◻BDEC), as required.