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ABCD is a rhombus with P,Q,R as mid points of AB,BC,CD respectively. Prove that PQ is perpendicular to QR

is any of the angle is given
Actually it has struck me already
The method to solve
My 16 points go in vain


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
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