• # question_answer ABCD is a parallelogram. BC is produced to Q such that BC = CQ. Then A)  area $\left( \Delta BCP \right)$ = area $\left( \Delta DPQ \right)$B)  area $\left( \Delta BCP \right)$ > area $\left( \Delta DPQ \right)$C)  area $\left( \Delta BCP \right)$ < area $\left( \Delta DPQ \right)$D)  area $\left( \Delta BCP \right)$ + area $\left( \Delta DPQ \right)$ = area $(\Delta \,BCD)$

(a): Join $AC\And DQ\therefore \Delta APC$ and $\Delta BCP$ lie on the same base PC and between the same parallels AB and PC $\therefore ar\left( \Delta APC \right)=ar\left( \Delta BCP \right)$             ....(i) Now, $AD\parallel CQ$ and $AD=CQ$ $\therefore$ $\Delta DQC$ is a parallelogram, Again $\Delta ADC$ and $\Delta DAQ$ are on the same base AD and between same parallels AD and CQ. $\therefore ar\left( \Delta ADC \right)=ar\left( \Delta ADQ \right)$ Subtracting ar (DAP) from both sides, we get ar$\left( \Delta APC \right)$=ar$\left( \Delta DPQ \right)$       .....(ii) From (i) and (ii), we get ar $\left( \Delta BCP \right)$ = ar$\left( \Delta DPQ \right)$
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