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
'.' each pair of opposite side is parallel
.'.ONQM is a parallelogram
angle MON = 90' (because diagonals of rhombus bisect each other at 90')
angle MON = angle PQR ( opposite angle of parallelogram)
.'. angle PQR = 90'