In the given figures, if PQRS is a parallelogram and `ABabs()PS`, then prove that `OCabs()SR`.

Explanation

Given PQRS is a parallelogram , so `PQabs()SR and PSabs()QR`. Also, `ABabs()PS`

To prove `OCabs()SR`
Proof in `DeltaOPS and DeltaOAB, PSabs()AB`
`anglePOS=angleAOB` [common angle]
`angleOSP=angleOBA` [corresponding angle]
`therefore DeltaOPS~DeltaOAB` [by AAA similarity criterion]
Then, `(PS)/(AB)=(OS)/(OB)`....(i)
In `DeltaCQR and DeltaCAB QRabs()PSabs()AB`
`angleQCR=angleACB` [common angle]
`angleCRQ=angleCBA` [corresponding angles]
`therefore DeltaCQR~DeltaCAB`
Then, `(QR)/(AB)=(CR)/(CB)`
`rArr (PS)/(AB)=(CR)/(CB)`
From Eqs. (i) and (ii), [since, PQRS is a parallelogram, so PS`equivQR`]
`(OS)/(OB)=(CR)/(CB)or(OB)/(OS)=(CB)/(CR)`
on subtracting from both sides, we get
`(OB)/(OS)-1=(CB)/(CR)-1`
`rArr (OB-OS)/(OS)=(CB-CR)/(CR)`
`rArr (BS)/(OS)=(BR)/(CR)`
By converse of basic proportionality theorem,
`SRabs()OC` Hence proved.