Prove that the quadrilateral formed by joining the midpoints of the sides of a rhombus is a rectangle.
Full process please
Answers
Step-by-step explanation:
Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a rectangle is a rhombus.
ANSWER
A rectangle ABCD in which P,Q,R and S are the mid-points of sides AB,BC,CD and DA respectively.
PQ,QR,RS and SP are joined.
Join AC. In ΔABC,P and Q are the mid-points of sides AB and BC respectively.
∴PQ∣∣AC and PQ=
2
1
AC .....(i) (Mid pt. Theorem)
In ΔADC,R and S are the mid-points of sides CD and AD respectively.
∴SR∣∣AC and SR=
2
1
AC ....(ii) (Mid. pt. Theorem)
From (i) and (ii), we have
PQ∣∣SR and PQ=SR
⇒PQRS is a parallelogram.
Now ABCD is a rectangle.
∴AD=BC
⇒
2
1
AD=
2
1
BC
⇒AS=BQ .....(iii)
In ΔS,APS and BPQ, we have
AP=BP (P is the mid-point of AB)
∠PAS=∠PBQ (Each is equal to 90
∘
)
and AS=BQ (From (iii))
∴ΔAPQ≅ΔBPQ (SAS)
⇒PS=PQ
∴PQRS is a parallelogram whose adjacent sides are equal.
⇒PQRS is a rhombus.
Answer: