### Theory:

Let us learn how to apply the theorems to solve the problems.
Example:
1. Show that a median divides a triangle into two triangles of equal area.

Solution: Given: $$PQR$$ is the triangle, and $$PS$$ is the median of the triangle.

To prove: $$ar(\triangle PQS) = ar(\triangle PRS)$$

Construction: Draw $$PX \perp BC$$

Proof: Consider $$\triangle PQS$$

$$ar(\triangle PQS) = \frac{1}{2} \times QS \times PX$$ ---- ($$1$$) [Area of a triangle is half the product of its base and the corresponding altitude.]

$$ar(\triangle PRS) = \frac{1}{2} \times SR \times PX$$ ---- ($$2$$) [Area of a triangle is half the product of its base and the corresponding altitude.]

From the figure, we can see that $$S$$ is the midpoint of $$QR$$.

Thus, $$QS = SR$$ ---- ($$3$$)

Substituting equation ($$3$$) in ($$1$$), we have:

$$ar(\triangle PQS) = \frac{1}{2} \times SR \times PX$$

$$ar(\triangle PQS) = ar(\triangle PRS)$$

Hence proved.

2. Diagonals $$PR$$ and $$QS$$ of a trapezium with $$PQ \parallel SR$$ intersect each other at $$O$$. Prove that $$ar(\triangle POS) = ar(\triangle QOR)$$.

Solution: Given: $$PQRS$$ is a trapezium with $$PQ \parallel SR$$. The diagonals $$PR$$ and $$QS$$ intersect at point $$O$$.

To prove: $$ar(\triangle POS) = ar(\triangle QOR)$$

Proof: The triangles $$PSR$$ and $$QSR$$ lie on the same base $$SR$$ and between the same parallels $$SR$$ and $$PQ$$.

Thus, $$ar(\triangle PSR) = ar(\triangle QSR)$$ [Triangles on the same base and between the same parallels are equal in the area.]

$$ar(\triangle PSR) - ar(\triangle SOR) = ar(\triangle QSR) - ar(\triangle SOR)$$ [Subtracting $$ar(\triangle SOR)$$ on both sides]

Therefore, $$ar(\triangle POS) = ar(\triangle QOR)$$

Hence proved.