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Answer :
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.