In triangle ABC, P and Q are the mid points of AB and BC. Thus PQ // AC (mid point theorem). We can also say it as NQ // MO ................... (i)
Similarly in triangle BCD, Q and R are the mid points. Thus, BD // QR. We can also say it as ON // MQ .................. (ii)

From (i) and (ii), we get that OMQN is a parallelogram. (NQ // MO and ON // MQ)
Now, we know that the diagonals of a rhombus bisect each other at 90°.
Thus, <BOC = 90°
Now as OMQN is a parallelogram, <BOC = <MQN = 90°

Hope that helps !!
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Let there be a rhombus ABCD.
to prove
draw the diagonals AC and BD.let them intersect at E
and it inersects PQ at F and QR at G
Consider the tri.BCD
llly, FQ//EG
i.e.EFGQ is a llgm
We know that diagonals of a rhombus are perpendicular bisectors of each other
thus,<FQG=90(opposite angles of a llgm are equal)
i.e.PQ is perpendicular to QR

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