In the figure AB - DC and AC and PQ intersect each other the point O- Prove that OA-CQ - OC-AP-

Question

Toppr Admin · Accepted Answer

In -x25B3-APO and -x25B3-CQO-x2220-AOP-x2220-COQ -vertically opposite angle-x2220-APO-x2220-CQO -alternative angle-x2220-PAO-x2220-QCO-xA0-alternative angle-Thus- OAAP-OCCQSo- OA-CQ-OC-APHence proved

In the figure $A B | | D C$ and $A C$ and $P Q$ intersect each other at the point $O$ . Prove that $O A . C Q = O C . A P$ .

In $△ A P O$ and $△ C Q O$
$∠ A O P = ∠ C O Q$ (vertically opposite angle)
$∠ A P O = ∠ C Q O$ (alternative angle)
$∠ P A O = ∠ Q C O$ (alternative angle)
Thus, $\frac{O A}{A P} = \frac{O C}{C Q}$
So, $O A . C Q = O C . A P$
Hence proved

In the figure AB||DC and AC and PQ intersect each other at the point O. Prove that OA.CQ=OC.AP.

In △APO and △CQO∠AOP=∠COQ (vertically opposite angle)∠APO=∠CQO (alternative angle)∠PAO=∠QCO (alternative angle)Thus, OAAP=OCCQSo, OA.CQ=OC.APHence proved

In the figure $A B | | D C$ and $A C$ and $P Q$ intersect each other at the point $O$ . Prove that $O A . C Q = O C . A P$ .

In $△ A P O$ and $△ C Q O$
$∠ A O P = ∠ C O Q$ (vertically opposite angle)
$∠ A P O = ∠ C Q O$ (alternative angle)
$∠ P A O = ∠ Q C O$ (alternative angle)
Thus, $\frac{O A}{A P} = \frac{O C}{C Q}$
So, $O A . C Q = O C . A P$
Hence proved