In this chapter we will prove that if any** parallelogram and triangle have same base and lie between same parallel line then the area of triangle is half of area of parallelogram**.

Consider the above figure in which;

ABCD is a parallelogram;

APD is a triangle;

Note that both parallelogram ABCD and triangle PAB;

-> have same base AB

-> lie between parallel lines AB and CP

**To Prove**:

Prove that area of triangle PAB is half of parallelogram ABCD;

i.e. **Area (Triangle PAB) = 1/2 ( Area of ABCD)**

**Proof**

First draw line BQ such that it’s parallel to line AP.

After joining BQ we get parallelogram ABQP.

Note that both parallelogram ABCD and ABQP lie in the same same AB and between parallel line AB & CP.

Using the parallelogram area theorem we can write;**Area (ABCD) = Area (ABQP)** —> eq (1)

Now consider parallelogram ABQP.

We know that in parallelogram, the diagonal divide the objects into two congruent triangle.

So in parallelogram ABQP, the diagonal BP divide it into two congruent triangles ABP and BPQ of equal area.

Area (APB) = Area (BPQ) = 1/2 Area (ABCD) **Area (ABCD)** = **2 * Area (APB)** = **2 * Area (BPQ)** —> eq (2)

Put the value of eq (2) in eq (1), we get;

2 * Area (APB) = Area (ABQP)

** Area (APB)** =** 1/2 Area (ABQP)**

Hence we proved that when parallelogram and triangle have same base and lie between same parallel lines then area of triangle is half of area of parallelogram.