Given AC and PQ intersect each other at the point O and AB II DC.
To prove that OA.CQ = OC. AP
Proof
In ΔAOP and ΔCOQ,
∠AOP=∠COQ [vertically opposite angles]
∠APO=∠CQO [since, AB II DC and PQ is transversal, so alternate angles]
∴ ΔAOP∼ΔCOQ [by AAA similarity criterion]
Then, OAOC=APCQ [since, corresponding sides are proportional]
⇒ OA.CQ = OC.AP