Prove that the parallelogram circumscribing a circle is a rhombus.

Or

Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

We have

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.