Prove a parallelogram circumscribing a circle as rhombus
Answers
Answer:
Step-by-step explanation:
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.
Hey Mate :D
Your Answer :---
Since ABCD is a parallelogram,
[ Plz see attached file also :) ]
AB = CD …(1)
BC = AD …(2)
It can be observed that
DR = DS (Tangents on the circle from point D)
CR = CQ (Tangents on the circle from point C)
BP = BQ (Tangents on the circle from point B)
AP = AS (Tangents on the circle from point A)
Adding all these equations, we obtain
DR + CR + BP + AP = DS + CQ + BQ + AS
(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)
CD + AB = AD + BC
On putting the values of equations (1) and (2) in this equation, we obtain
2AB = 2BC
AB = BC …(3)
Comparing equations (1), (2), and (3), we obtain
AB = BC = CD = DA
Hence, ABCD is a rhombus.
Like this answer ? Hit like button n Follow me ❤
Glad To Help :D
