Question

In the given figure, $AP\parallel BQ\parallel CR$. Prove that area(ACQ) is equal to area(PBR).

Answer 1

Hint:
Two figures are said to be on the same base and between same parallels, if they have a common base (side) and the vertices, (or the vertex) opposite to the common base of each figure lie on a line parallel to the base.
The following theorem can be useful in dealing with these types of questions.
 Theorem 1: Two triangles on the same base (or equal bases) and between the same parallel lines are equal in area.

Complete step by step solution:
Step 1: Draw the given figure

Step 2
Given that $AP\parallel BQ$ …… (1)
Consider $\vartriangle ABQ$and $\vartriangle PQB$ , having the same base BQ and are between the parallels AP and BQ
$ \Rightarrow ar(\vartriangle ABQ) = ar(\vartriangle PQB)$ (according to Theorem 1) …… (2)

Step 3
Given that $BQ\parallel CR$ …… (3)
Consider $\vartriangle CBQ$and $\vartriangle RQB$ , having the same base BQ and are between the parallels BQ and CR

$ \Rightarrow ar(\vartriangle CBQ) = ar(\vartriangle RQB)$ (according to theorem 1) …… (4)
Step 4: Adding (2) and (4)
$
   \Rightarrow ar(\vartriangle ABQ) + ar(\vartriangle CBQ) = ar(\vartriangle PQB) + ar(\vartriangle RQB) \\
   \Rightarrow {\text{ }}ar(\vartriangle AQC) = ar(\vartriangle PBR) \\
 $
Hence proved. The required result area(ACQ) is equal to area(PBR) has been proved.

Note:
The converse (or reverse) of the above mention theorem is also true, i.e.
Theorem 2: Two triangles having the same base (or equal bases) and given that their areas are equal; then they lie between the same parallel lines.
Theorem 1 and 2 are also true for the parallelograms.
We should not get confused with complex figures, try to break it in smaller areas.
For the application of theorem, look for parallel lines and triangles or parallelograms between them.