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D and E are the mid-points of the sides AB and AC of DABC and O is any point on side BC. O is joined to A. If P and
Q are the mid-points of OB and OC respectively, 
then DEQP is
(A) a square                  (B) a rectangle
(C) a rhombus              (D) a parallelogram

Answers (1)

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Answer: (D) a parallelogram
Solution.

By midpoint theorem
DE\parallel BC                    …..(1)
i.e.,  DE=\frac{1}{2}(BC)
DE=\frac{1}{2}(BP+PO+OQ+QC)
DE=\frac{1}{2}(2PO+2OQ)
{\because P and Q are midpoints of OB and OC}
DE=PO+OQ
DE=PQ                           …..(2)
Now in \triangle AOC, Q and E are the mid points of OC and AC.
\therefore EQ\parallel AO and EQ=\frac{1}{2}AO                …..(3)
{Using mid-point theorem}
Similarly, in   \triangle ABO,PD\parallel AO  and  PD=\frac{1}{2}AO          …..(4)
From equation 3 and 4
EQ\parallel PD and  EQ=PD
From equation 1 and 3
DE\parallel PQ and  DE= PQ

Hence DEQP is a parallelogram.
Therefore option (D) is correct.

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