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:
 
17.png
 
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:
 
18.png
 
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.