Question

In figure, if PQRS is a parallelogram and AB || PS, then prove that OC || SR.

Answer 1

Given,

PQRS is a parallelogram,

Therefore, PQ || SR and PS || QR.

Also given, AB || PS.

To prove:

OC || SR

From ∆OPS and OAB,

PS||AB

∠POS = ∠AOB ......[Common angle]

∠OSP = ∠OBA   ......[Corresponding angles]

∆OPS ∼ ∆OAB   ......[By AAA similarity criteria]

Then,

Using basic proportionality theorem,

We get,

`(PS)/(AB) = (OS)/(OB)`   ......(i)

From ∆CQR and ∆CAB,

QR || PS || AB

∠QCR = ∠ACB   ......[Common angle]

∠CRQ = ∠CBA   .......[Corresponding angles]

∆CQR ∼ ∆CAB

Then, by basic proportionality theorem

`(QR)/(AB) = (CR)/(CB)`

`(PC)/(AB) = (CR)/(CB)`  ......(ii)

[PS ≅ QR Since, PQRS is a parallelogram,]

From Equation (i) and (ii),

`(OS)/(OB) = (CR)/(CB)`

`(OB)/(OS) = (CB)/(CR)`

Subtracting 1 from L.H.S and R.H.S, we get,

`(OB)/(OS) - 1 = (CB)/(CR) - 1`

`(OB - OS)/(OS) = ((CB - CR))/(CR)`

`(BS)/(OS) = (BR)/(CR)`

SR || OC  ......[By converse of basic proportionality theorem]

Hence proved.