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# In Figure, ar (DRC) = ar (DPC) and ar (BDP) = ar (ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums.

Given:
Area ($$\triangle{DRC}$$) = Area ($$\triangle{DPC}$$).

As, $$\triangle{DRC}$$ and $$\triangle{DPC}$$ lie on the same base DC and have equal area, therefore, they must lie between the same parallel lines.

$$\because$$ DC || RP
$$\therefore$$ DCPR is a trapezium.

Also, Area ($$\triangle{BDP}$$) = Area ($$\triangle{ARC}$$).

$$\Rightarrow$$ Area ($$\triangle{BDP}$$) - Area ($$\triangle{DPC}$$) = Area ($$\triangle{ARC}$$) - Area ($$\triangle{DRC}$$)

$$\therefore$$ Area ($$\triangle{BDC}$$) = Area($$\triangle{ADC}$$)

$$\because$$ $$\triangle{BDC}$$ and $$\triangle{ADC}$$ lie on the same base CD and have equal areas, they must lie between the same parallel lines.

$$\therefore$$ AB || CD
Hence, it is proved that ABCD is a trapezium.