In the given figure, if PQRS …

To prove - OC║SR

Proof  -  In ΔOPS and ΔOAB

∠POS = ∠AOB     (common in both)

∠OSP = ∠OBA  (corresponding angles are equal as PS║AB)

=> ΔOPS ~ ΔOAB    [AA criteria]

=>  PS/AB =  OS/OB       ........................(1)     (sides in similar triangles are proportional)

In ΔCAB and ΔCRQ

As, QR║AB

=> ∠QCR = ∠ACB        (common)

=>  ∠CBA  = ∠CRQ       (corresponding angles are equal)

=>  ΔCAB ~ ΔCQR           [AA criteria]

=>  CR/CB = QR/AB         (sides in similar triangles are proportional)

Also,  PS = QR        [ PQRS  is parallelogram]

=>  CR/CB = PS/AB                     ......................(2)

From    (1) and (2)

=>   OS/OB =  CR/CB

=>   OB/OS =  CB/CR

Subtracting 1 from both sides

So,  OB/OS - 1 =  CB/CR - 1

=>  (OB - OS)/OS  =  (CB - CR)/CR

=>  BS/OS = BR/CR

By converse of Thales Theorem

=>  OC║SR        .     Hence proved  .