Given: ABCD is a trapezium. AB∥CD, P is mid point of AC and Q is mid point of BD
Now, In △CDQ and △RBQ
∠QDC=∠QBR .....(Alternate angles)
∠CQD=∠BQR ....(Vertically Opposite angles)
DQ=BQ .....(Q is mid point of BD)
Thus, △DQC≅△BQR ....(ASA rule)
Hence, by CPCT, CQ=QR and DC=RB
Now, In ΔACR, P is mid-point of AC and Q is mid-point of RC
By Mid Point Theorem, we have
PQ=12AR
PQ=12(AB−BR)
PQ=12(AB−DC) ....(Since, DC=RB)