if AB || DC and AC and PQ intersect each other at the point O, prove that OA × CQ = OC × AP
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Given AC and PQ intersect each other at the point O and AB || DC
Prove that OA × CQ = OC × AP
Proof in $\triangle\text{AOP}$ and $\triangle\text{COQ},$
$\angle\text{AOP}=\angle\text{COQ}$ [vertically opposite angles]
$\angle\text{APO}=\angle\text{COQ}$
[Since, AB || DC and PQ is transversal, so alternate angles]
$\therefore\triangle\text{AOP}\sim\triangle\text{COQ}$ [by AAA similarity citerion]
Then $\frac{\text{OA}}{\text{OC}}=\frac{\text{AP}}{\text{CQ}}$ [Since, corresponding sides are proportional]
⇒ OA × CQ = OC × AP Hence proved.
Prove that OA × CQ = OC × AP
Proof in $\triangle\text{AOP}$ and $\triangle\text{COQ},$
$\angle\text{AOP}=\angle\text{COQ}$ [vertically opposite angles]
$\angle\text{APO}=\angle\text{COQ}$
[Since, AB || DC and PQ is transversal, so alternate angles]
$\therefore\triangle\text{AOP}\sim\triangle\text{COQ}$ [by AAA similarity citerion]
Then $\frac{\text{OA}}{\text{OC}}=\frac{\text{AP}}{\text{CQ}}$ [Since, corresponding sides are proportional]
⇒ OA × CQ = OC × AP Hence proved.
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