Given: ABCD is a rhombus

P, Q, R are then mid points of AB, BC, CD.

to proove: angle PQR = 90

Construction: Join AC and BD.

Proof : In ΔDBC

R and Q are the mid points of DC and CB

.'.RQ // DB ( by mid point theorem)

.'. MQ//ON ..............eq.1 (parts of RQ and DB)

Now , in Δ ACB

P and Q are the mid points of AB and BC

.'. AC // PQ ( by mid point theorem)

.'. OM //NQ .................eq.2 ( OM and NQ are the parts of AC and PQ)

from eq. 1 and 2

MQ //ON

ON //NQ

'.' each pair of opposite side is parallel

.'.ONQM is a parallelogram

in ONQM

angle MON = 90' (because diagonals of rhombus bisect each other at 90')

angle MON = angle PQR ( opposite angle of parallelogram)

__.'. angle PQR = 90'__