### Theory:

Let us learn how to apply the theorem to the given problem to find the solution.
Example:
If a triangle and a parallelogram are on the same base and between the same parallels, then prove that the area of a triangle is equal to half the area of the parallelogram.

Solution:

Given: $$PQRS$$ is a parallelogram, and $$OSR$$ is a triangle. They both are on the same base $$SR$$ and between the same parallels $$SR$$ and $$OQ$$.

To prove: $$ar(\triangle OSR) = \frac{1}{2} \times ar(PQRS)$$

Construction: Draw $$OX \perp SR$$ and $$PY \perp SR$$

Proof: Consider the parallelogram $$PQRS$$.

$$ar(PQRS) = PY \times SR$$ ---- ($$1$$)

Similarly, consider the triangle $$ORS$$.

$$ar(ORS) = \frac{1}{2} \times OX \times SR$$ ---- ($$2$$)

In parallelogram $$PQRS$$, $$PQ \parallel RS$$ [Opposite sides of a parallelogram are parallel]

Thus, $$OX = PY$$ ---- ($$3$$) [Distance between parallel lines are equal]

Substituting equation ($$3$$) in ($$2$$, we get:

$$ar(ORS) = \frac{1}{2} \times PY \times SR$$

$$ar(ORS) = \frac{1}{2} \times ar(PQRS)$$ [Using ($$1$$)]

Hence proved.