can someone explain this plz.........
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Step-by-step explanation:
Let ABCD be a rhombus and P,Q,R and S be the mid-points of sides AB,BC,CD and DA respectively.
In △ABD and △BDC we have
SP∥BD and SP=
2
1
BD ---- ( 1 ) [ By mid-point theorem ]
RQ∥BD and RQ=
2
1
BD ---- ( 2 ) [ By mid-point theorem ]
From ( 1 ) and ( 2 ) we get,
SP∥RQ
PQRS is a parallelogram
As diagonals of a rhombus bisect each other at right angles.
∴ AC⊥BD
Since, SP∥BD,PQ∥AC and AC⊥BD
∴ SP⊥PQ
∴ ∠QPS=90
o
∴ PQRS is a rectangle.
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