Question

PT opposite sides of a quadrilateral circumscribing a circle sutent supplementary angle at the centre of the circle

Answer 1

you also go for diagram given

Given: ABCD is a parallelogram which touches a circle at P, Q, R and S.
To Prove: ABCD is a rhombus, i.e., AB = BC = CD = DA
Proof: We know, AB = CD & AD = BC [∵, ABCD is a parallelogram and opposite sides of parallelogram are equal] …(i)
Now AB = AP + PB [clearly from the diagram]
Or AB = AS + BQ [∵, AP = AQ & PB = BQ, as tangents from an external point are equal in length] …(ii)
Also, CD = CR + RD [from the diagram]
Or CD = CQ + SD [∵, CR = CQ & RD = SD, as tangents from an external point are equal in length] …(iii)
Adding equation (i) and (ii),
AB + CD = AS + BQ + CQ + SD
As AB = CD from equation (i), we can write
AB + AB = (AS + SD) + (BQ + CQ)
⇒ 2AB = AD + BC
⇒ 2AB = AD + AD [∵, from equation (i)]
⇒ 2AB = 2AD
⇒ AB = AD
From equation (i), AB = CD and AD = DA
And AB = AD
Thus, AB = BC = CD = DA
Hence, proved.
Or
We have,
To prove: ∠AOD + ∠BOC = 180°
And ∠AOB + ∠COD = 180°
Proof: Taking ∆BPO & ∆BQO,
We can see
BP = BQ [∵, tangents from an external point are of equal length]
PO = OQ [∵, radii of a circle are always equal]
BO = BO [∵, common]
⇒ [by sss congruency rule]
Thus by similar triangle’s property, ∠8 = ∠1 …(i)
Similarly for other triangles,
∠7 = ∠6 …(ii)
∠5 = ∠4 …(iii)
∠3 = ∠2 …(iv)
Adding all these angles of the circle, we get
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360° [∵, summation of all angles of a circle beginning and ending at the same point is 360°]
⇒ ∠1 + ∠2 + ∠2 + ∠4 + ∠4 + ∠6 + ∠6 + ∠1 = 360° [from equations (i), (ii), (iii) and (iv)]
⇒ 2 (∠1 + ∠2 + ∠4 + ∠6) = 360°
⇒ (∠1 + ∠2) + (∠4 + ∠6) = 180°
⇒ ∠BOC + ∠DOA = 180° [∵, (∠1 + ∠2) = ∠BOC & (∠4 + ∠6) = ∠AOD]
Also, ∠AOB + ∠BOC + ∠COD + ∠DOA = 360°
⇒ ∠AOB + ∠COD + (∠BOC + ∠DOA) = 360°
⇒ ∠AOB + ∠COD + 180° = 360°
⇒ ∠AOB + ∠COD = 360° - 180°
⇒ ∠AOB + ∠COD = 180°
Thus, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Hence, proved.

Answer 2

Hey Mate :D

Your Answer :---

Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle at point P, Q, R, S.

Let us join the vertices of the quadrilateral ABCD to the center of the circle.

[ plz see attached file also :) ]

Consider ∆OAP and ∆OAS,

AP = AS (Tangents from the same point)

OP = OS (Radii of the same circle)

OA = OA (Common side)

∆OAP ≅ ∆OAS (SSS congruence criterion)

Therefore, A ↔ A, P ↔ S, O ↔ O

And thus,

∠POA = ∠AOS ∠1 = ∠8 Similarly, ∠2 = ∠3 ∠4 = ∠5 ∠6 = ∠7

∠1 + ∠2 + ∠ 3 + ∠4 + ∠5 + ∠ 6 + ∠7 + ∠ 8 = 360º

( ∠ 1 + ∠ 8) + ( ∠ 2 + ∠ 3) + ( ∠4 + ∠ 5) + ( ∠6 + ∠ 7) = 360º

2 ∠1 + 2 ∠2 + 2∠ 5 + 2∠ 6 = 360º

2( ∠1 + ∠ 2) + 2( ∠5 + ∠ 6) = 360º

( ∠1 + ∠2 ) + ( ∠5 + ∠ 6) = 180º

∠AOB + ∠COD = 180º

Similarly, we can prove that

∠BOC + ∠DOA = 180º

Hence, opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

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