# The figure formed by joining the mid-points of the sides of a quadrilateral ABCD, taken in order, is a square only if,A. ABCD is a rhombusB. diagonals of ABCD are equalC. diagonals of ABCD are equal and perpendicularD. diagonals of ABCD are perpendicular.

The Midpoint Theorem states that the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.

PQRS is a square

PQ = QR = RS = SP

Also, PR = SQ [Diagonals of a square]

And,

BC = PR

SQ = AB

BC = PR = SQ = AB

i.e, BC = AB

Thus all sides of ABCD are equal

ABCD is a square or a rhombus.

In ∆ABD,

S and P are midpoints of AD and AB

By midpoint theorem,

SP DB and SP = 1/2DB

Similarly,

In ∆ABC,

Q and P are midpoints of BC and AB

By midpoint theorem,

PQ AC and PQ = 1/2AC

PQRS is a square

SP = PQ

1/2DB = 1/2AC

DB = AC

i.e, Diagonals of ABCD are equal

it cannot be a square

Hence, it is a rhombus

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