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:
 
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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.