# Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.

**Solution:**

Let's draw the given quadrilateral ABCD.

It is given that,

Area (ΔAOD) = Area (ΔBOC)

Now, adding Area (ΔAOB) on both the sides

area (ΔAOB) + area (ΔAOD) = area (ΔAOB) + area (ΔBOC)

Hence ar (ΔADB) = ar (ΔACB)

According to Theorem 9.3: Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.

Therefore, ΔADB and ΔACB, are lying between the parallels lines. i.e., AB || CD as they lie on the same base AB and have equal areas.

Therefore, ABCD is a trapezium.

**Video Solution:**

## Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.

### Maths NCERT Solutions Class 9 - Chapter 9 Exercise 9.3 Question 15:

**Summary:**

If diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC), then ABCD is a trapezium.