The Brainliest Answer!
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'

2 5 2
Yes it is correct for sure
N is the point wherePQ intersect BD
O is where diagonals intersect
and M is the point where RQ intersect AC.
I had drawn its figure but forgot to add it with question