Question

can someone explain this plz.........​

Answer 1

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.