# Q : 3      P and Q are any two points lying on the sides DC and AD respectively of a parallelogram ABCD. Show that $\small ar(APB)=ar(BCQ)$.

We have,
ABCD is a parallelogram, therefore AB || CD and BC || AD.

Now,  $\Delta$APB and ||gm ABCD are on the same base AB and between two parallels AB and DC.
Therefore, ar  ($\Delta$APB) = 1/2 . ar(||gm ABCD)...........(i)

Also, $\Delta$BQC and ||gm ABCD are on the same base BC and between two parallels BC and AD.
Therefore, ar($\Delta$BQC) = 1/2 . ar(||gmABCD)...........(ii)

From eq(i) and eq (ii),  we get,
ar  ($\Delta$APB) =  ar($\Delta$BQC)
Hence proved.

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