Prove that the parallelogram circumscribing a circle is a 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.
Answer:
Step-by-step explanation:
the solution is given below
Given :- circle with Centre O and also given ABCD is the llgm
We need to prove that ABCD is a rhombus
pls mark the pt of contact For AB - P BC- Q CD-R DA-S Proof:- Tangents from an external point to a circle are equal in length.
Therfore AP=AS
BP=BQ
DR=DS
CR=CQ
Adding all the LHS and RHS we get
AB+CD=AD+BC eq.1
Since ABCD is llgm
Opp sides will be equal. Therefore AB=CD & AD=BC
when u substitute the above in eq.1 we get 2AB=2AD
=> AB=AD
=> AB=BC=CD=DA
Therefore ABCD is a rhombus