Question

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


Any answer???
pls do this favour!!​

Answer 1

Answer:

here AB=CD and BC=DA

Step-by-step explanation:

and PQRS are apoints

Answer 2

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= ½AC

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= ½AC

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.