Question

ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.​

Answer 1

""" ❤️ Answer ❤️ """

Here, we are joining A and C.

In ΔABC

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

=

2

1

AC

In ΔADC

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

=

2

1

AC

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

In ΔAPS & ΔBPQ

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

∠ PAS

=

∠

PBQ(All the angles of rectangle are 90

o

)

AS=BQ

≅

Δ

∴ Δ

∴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.