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(i) In ΔDAC,
R is the mid point of DC and S is the mid point of DA.
Thus by mid point theorem, SR || AC and SR = ½ AC
(ii) In ΔBAC,
P is the mid point of AB and Q is the mid point of BC.
Thus by mid point theorem, PQ || AC and PQ = ½ AC
also, SR = ½ AC
, PQ = SR
(iii) SR || AC ———————- from question (i)
and, PQ || AC ———————- from question (ii)
⇒ SR || PQ – from (i) and (ii)
also, PQ = SR
, PQRS is a parallelogram
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