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Updated on : 2022-09-05

Solution

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Here, we are joining A and C.

In $Δ$ABC

P is the mid point of AB

Q is the mid point of BC

PQ$∣∣$AC [Line segments joining the mid points of two sides of a triangle is parallel to AC(third side) and also is half of it]

PQ$=21 $AC

In $Δ$ADC

R is mid point of CD

S is mid point of AD

RS$∣∣$AC [Line segments joining the mid points of two sides of a triangle is parallel to third side and also is half of it]

RS$=21 $AC

So, PQ$∣∣$RS and PQ$=$RS [one pair of opposite side is parallel and equal]

In $Δ$APS & $Δ$BPQ

AP$=$BP [P is the mid point of AB)

$∠$PAS$=∠$PBQ(All the angles of rectangle are $90_{o}$)

AS$=$BQ

$∴Δ$APS$≅Δ$BPQ(SAS congruency)

$∴$PS$=$PQ

BS$=$PQ & PQ$=$RS (opposite sides of parallelogram is equal)

$∴$ PQ$=$RS$=$PS$=$RQ[All sides are equal]

$∴$ PQRS is a parallelogram with all sides equal

$∴$ So PQRS is a rhombus.

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