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# 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.