In ΔABC,
P is mid-point to AB [∵ Given]
Q is mid-point of BC [∵ Given]
Hence, PQ ∥ AC and PQ = (1/2)AC … (1) [∵ Mid-point Theorem]
Similarly, in ΔACD,
S is mid-point of AD [∵ Given]
R is mid-point of CD [∵ Given]
Hence, SR ∥ AC and SR = (1/2) AC …(2) [∵ Mid-Point Theorem]
From (1) and (2), we have
PQ ∥ SR …(3) [∵ PQ ∥ AC and SR ∥ AC]
and PQ = SR …(4) [∵ SR = (1/2)AC and PQ = (1/2)AC]
Hence, PQRS is a parallelogram.
Similarly, in ΔBCD,
Q is mid-point of BC [∵ Given]
R is mid-point of CD [∵ Given]
Hence, QR ∥ BD [∵ Mid-point Theorem]
⇒ QN ∥ LM …(5)
and LQ ∥ MN …(6) [∵ PQ ∥ AC]
From (5) and (6), we have
LMNQ is a parallelogram.
Hence, ∠LMN = ∠LQN [∵ Opposite angles of a parallelogram]
But, ∠LMN = 90° [∵ Digonals of a rhombus are perpendicular to each other]
Hence, ∠LQN = 90°
A parallelogram whose one angle is a rectangle. Hence, PQRS is a rectangle.