Question

Given a parallelogram ABCD where P, Q, R and S are the midpoints of AB, BC, CD and DA respectively. Show that SP is parallel to QR and PQ is parallel to SR. plz answer it's important

Answer 1

Answer:

Given, ABCD is rhombus, where P,Q,R and S are the mid points of AB,BC,CD and DA respectively.

We have to prove PQRS is a rectangle

First we will prove PQRS is parallelogram, since parallelogram with one angle 90

o

is rectangle.

In ΔABC, P is mid point of AB, Q is mid point of BC.

∴PQ∣∣AC and PQ=

2

1

AC..(1)

In ΔADC, R is the mid-point of CD, S is mid point AD.

∴RS∣∣AC and RS=

2

1

AC (2)

From (1) and (2)

In PQRS, one pair of opposite side is parallel and equal.

Hence, PQRS is a parallelogram.

Now we have to prove PQRS is a rectangle

Since AB=BC⇒

2

1

AB=

2

1

BC

So PB=BQ

Now, In ΔBPQ, PB=BQ

∴∠2=∠1 (3)

In ΔAPS & ΔCQR

⇒AP=CQ

⇒AS=CR

⇒PS=QR

∴ΔAPS≅ΔCQR

⇒∠3=∠4 by CPCT …(4)

Now

AB is a line

So, ∠3+∠SPQ1=180

o

.(5)

Similarly for line Bc

∠2+∠PQR+∠4=180

o

∠1+∠PQR+∠3=180

o

6)

From 5 & 6

∠1+∠SPQ+∠3=∠1+∠PQR+∠3

∴∠SPQ=∠PQR.(7)

Now, PS=QR, and PQ is a transversal

So, ∠ SPQ+∠PQR=180

o

∠SPQ+∠SPQ=180 from 7

2∠SPQ=180

o

∠SPQ=90

o

So, PQRS is a parallelogram with one angle 90

o

∴ PQRS is a rectangle.