In the given figure, AP ∥ BQ ∥ CR. What will be (△AQC)?
ar (△PBR)
Since △ABQ and △PBQ lie on the same base BQ and are between the same parallels AP and BQ,
∴ Area (△ABQ) = Area (△PBQ) ... (1)
Again, △BCQ and △BRQ lie on the same base BQ and are between the same parallels BQ and CR.
∴ Area (△BCQ) = Area (△BRQ) ... (2)
On adding equations (1) and (2), we obtain
Area (△ABQ) + Area (△BCQ) = Area (△PBQ) + Area (△BRQ)
⇒ Area (△AQC) = Area (△PBR)