Theory:

Theorem I: Area of a triangle is equal to half the product of its base and the corresponding altitude.
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Given: \(PQRS\) is a parallelogram, and \(PR\) is one of its diagonal.
 
Construction: Draw \(PX \perp SR\).
 
To prove: \(ar(\triangle PSR) = \frac{1}{2} \times SR \times PX\)
 
Proof: Since \(\triangle PRS\) and parallelogram \(PQRS\) are on the same \(SR\) and between the same parallels, then we have:
 
\(ar(PRS) = \frac{1}{2} \times ar(PQRS)\)
 
\(ar(PRS) = \frac{1}{2} \times SR \times PX\) [Area of a parallelogram \(=\) base \(\times\) height]
 
Therefore, the area of a triangle is equal to half the product of its base and the corresponding altitude.
 
Hence proved.
Theorem II: Two triangles on the same base and between the same parallels are equal in area.
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Given: \(\triangle ABC\) and \(\triangle DBC\) are on the same base \(BC\) and between the same parallels \(BC\) and \(XY\).
 
To prove: \(ar(\triangle ABC) = ar(\triangle DBC)\)
 
Construction: Draw \(BP \parallel AC\) and \(CQ \parallel BD\) meeting \(XY\) at \(P\) and \(Q\).
 
Proof: From the figure, we can see that \(BCAP\) and \(BCQD\) are parallelogram and are on the same base \(BC\) and between the same parallels \(BC\) and \(XY\).
 
\(ar(BCAP) = ar(BCQD)\) ---- (\(1\)) [Parallelograms on the same base and between the same parallels are equal in the area]
 
Also, \(\triangle ABC\) and parallelogram \(BCAP\) are on the same base \(BC\) and between the same parallels \(BC\) and \(XY\).
 
\(ar(\triangle ABC) = \frac{1}{2} \times ar(BCAP)\) ---- (\(2\)) [If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of a parallelogram]
 
Similarly, \(\triangle DBC\) and parallelogram \(BCQD\) are on the same base \(BC\) and between the same parallels \(BC\) and \(XY\).
 
\(ar(\triangle DBC) = \frac{1}{2} \times ar(BCQD)\) ---- (\(3\)) [If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is equal to half the area of a parallelogram]
 
Substituting the value of equations (\(2\)) and (\(3\)) in (\(1\)), we have:
 
\(2 \times ar(\triangle ABC) = 2 \times ar(\triangle DBC)\)
 
\(ar(\triangle ABC) = ar(\triangle DBC)\)
 
Hence proved.
Theorem III: Two triangles having the same base and equal areas lie between the same parallels.
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Given: \(PRS\) and \(QRS\) are two triangles on the same base \(SR\) with equal areas.
 
That is, \(ar(\triangle PRS) = ar(\triangle QRS)\)
 
To prove: The two triangles lie between the same parallels. That is, \(PQ \parallel SR\)
 
Construction: Join \(PQ\). Draw \(PX \perp SR\) and \(QY \perp SR\)
 
Proof: Since \(PX \perp SR\) and \(QY \perp SR\), then
 
\(PX \parallel QY\) ---- (\(1\)) [Lines perpendicular to the same line are parallel to each other]
 
Also, we are given that \(ar(\triangle PRS) = ar(\triangle QRS)\)
 
\(\frac{1}{2} \times RS \times PX = \frac{1}{2} \times RS \times QY\)
 
\(PX = QY\) ---- (\(2\))
 
Now, consider \(PQXY\),
 
\(PX \parallel QY\) and \(PX = QY\) [From equations (\(1\)) and (\(2\))]
 
Therefore, \(PQXY\) is a parallelogram. Because a pair of opposite sides are equal and parallel.
 
Thus, \(PQ \parallel XY\) [Opposite sides of a parallelogram are parallel]
 
\(PQ \parallel RS\) [Since \(XY\) lies on the line \(RS\)]
 
Therefore, two triangles having the same base and equal areas lie between the same parallels.
 
Hence proved.