Question

In the adjoining figure, O is the mid-point of diagonal AC of a quadrilateral ABCD. Prove that area (◻ABOD) = area (◻BODC).

Answer 1

Given: O is the midpoint of diagonal AC of quadrilateral ABCD, so AO = OC. Points A, O and C are collinear.

To Prove: Area (ABOD) = Area (BODC).

Proof [Step-wise]:

1. Split the quadrilaterals by the diagonals through O:

Quadrilateral ABOD = area (ΔAOB) + area (ΔAOD) split by AO.

Quadrilateral BODC = area (ΔBOC) + area (ΔCOD) split by OC.

2. Compare ΔAOB and ΔBOC:

Both triangles have the same vertex B and their bases AO and OC lie on the same line AC, so the perpendicular distance (height) from B to line AC is the same for both triangles.

Area (ΔAOB) = `1/2` × AO × height from B to AC.

And Area (ΔBOC) = `1/2` × OC × height from B to AC

Since AO = OC   ...(O is midpoint of AC), area (ΔAOB) = area (ΔBOC)

3. Compare ΔAOD and ΔCOD similarly:

Both have vertex D and bases AO and OC on the same line AC, so the perpendicular distance from D to AC is the same for both.

Area (ΔAOD) = `1/2` × AO × height from D to AC.

And Area (ΔCOD) = `1/2` × OC × height from D to AC.

Again, AO = OC, so area (ΔAOD) = area (ΔCOD).

4. Add the equalities from steps 2 and 3:

Area (ΔAOB) + Area (ΔAOD) = Area (ΔBOC) + Area (ΔCOD).

5. Using step 1, the left side is area (ABOD) and the right side is area (BODC). 

Therefore, area (ABOD) = area (BODC).

Hence proved.