prove that two triangles having the same base (or equal bases) and equal areas lie between the same parallels.

**Given: **Δ ABC and Δ PBC lie on the same base BC and having equal areas.

**To prove:** ΔABC and ΔPBC lies between the same parallels *l* and *m*.

**Construction:** Draw AD and PQ, the altitude drawn from A and P on BC respectively.

**Proof :**

Given, ar (ΔABC) = ar(ΔPBC)

We know that area of a triangle

Therefore, Area (ΔABC)

and Area (ΔPBC)

On equating (1) and (2), we get

⇒ The height of perpendiculars drawn between two lines *l* and *m* are equal which is possible only in one case when the lines *l* and *m* are running parallel to each other.

Hence, the triangles ABC and PBC drawn on the same base BC and having equal areas lying between the same parallels *l* and *m*.

[Hence Proved]

**
**