Question

ABCD is a rectangle and P, Q, R, and S are midpoints of the sides. AB, BC, CD and DA respectively show that the quadrilateral ​

Answer 1

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=  1 /2AC

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 )

AS=BQ

∴ΔAPS≅ΔBPQ(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.