In Figure, AP || BQ || CR. Prove that ar (AQC) = ar (PBR).

It can be observed that, \(\triangle{ABQ}\) and \(\triangle{PBQ}\) lie on the same base BQ and are between the same parallels AP and BQ.

\(\therefore \) Area (\(\triangle{ABQ}\)) = Area (\(\triangle{PBQ}\)) ...(i)

Also, \(\triangle{BCQ}\) and \(\triangle{BRQ}\) lie on the same BQ and are between the same parallels BQ and CR.

\(\therefore \) , Area (\(\triangle{BCQ}\)) = Arca (\(\triangle{BRQ}\)) ...(ii)

From Equations (i) and (ii), we get,
\(\Rightarrow \) Area (\(\triangle{ABQ}\)) + Area(\(\triangle{BCQ}\)) = Area (\(\triangle{PBQ}\)) + Area (\(\triangle{BRQ}\))

\(\Rightarrow \) Area (\(\triangle{AQC}\)) = Area (\(\triangle{PBR}\))
Hence, proved.