Question

5F
Q16. ABCD is a rectangle and
P, Q, R and S are mid-points of
the sides AB, BC, CD and DA
respectively. Show that the
quadrilateral PQRS is a
rhombus. OR In the figure
given below the side QR of A
PQR is produced to a point S.
If the bisectors of _ PQR and 2
PRS meet at point T, then
prove that < QTR = 12 _ QPR​

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

∴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✿◉●•◦