Prove that the parallelogram circumscribing a circle is a rhombus.
Answers
Answer:
Given ABCD is a ||gm such that its sides touch a circle with centre O.
∴ AB = CD and AB || CD,
AD = BC and AD || BC
Now, P, Q, R and S are the touching point of both the circle and the ||gm
We know that, tangents to a circle from an exterior point are equal in length.
∴ AP = AS [Tangents from point A] ... (1)
BP = BQ [Tangents from point B] ... (2)
CR = CQ [Tangents from point C] ... (3)
DR = DS [Tangents from point D] ... (4)
On adding (1), (2), (3) and (4), we get
AP + BP + CR + DR = AS + BQ + CQ + DS
⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)
⇒ AB + CD = AD + BC
⇒ AB + AB = BC + BC [∵ ABCD is a ||gm . ∴ AB = CD and AD = BC]
⇒ 2AB = 2BC
⇒ AB = BC
Therefore, AB = BC implies
AB = BC = CD = AD
Hence, ABCD is a rhombus.
In rhombus, it is not necessary that diagonals are equal. If they are equal, then rhombus is considered as a square whose diagonals are always equal. So, there isn't any use of proving that the diagonals of a rhombus are equal
Consider a circle circumscribed by a parallelogram ABCD, Let side AB, BC, CD and AD touch circles at P, Q, R and S respectively.
To Proof : ABCD is a rhombus.
As ABCD is a parallelogram
AB = CD and BC = AD …[1]
[opposite sides of a parallelogram are equal]
Now, As tangents drawn from an external point are equal.
We have
AP = AS
[tangents from point A]
BP = BQ
[tangents from point B]
CR = CQ
[tangents from point C]
DR = DS
[tangents from point D]
Add the above equations
AP + BP + CR + DR = AS + BQ + CQ + DS
AB + CD = AS + DS + BQ + CQ
AB + CD = AD + BC
AB + AB = BC + BC [From 1]
AB = BC …[2]
From [1] and [2]
AB = BC = CD = AD
And we know,
A parallelogram with all sides equal is a rhombus
So, ABCD is a rhombus.
Hence Proved.