In the figure, $P$ is the centre of the circle. Two chords $A B$ and $C D$ are parallel to each other. Prove $∠ C P A = ∠ D P B$ .

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Solution

Consider $Δ O P C$ and $Δ O P D$
Since, $O P = O P$ as a common side.
$∠ P O C = P O D$ [Right angle]
$P C = P D$
Thus, $Δ P O C ≅ Δ P O D$ from RHS Criterion
So, $∠ O P C = ∠ O P D$ .....(i)
Similar way, we can prove $Δ L P A ≅ Δ L P B$
So, $∠ L P A = ∠ L P B$ .....(ii)
From equation(ii)-equation(i), we get
$\Rightarrow ∠ L P A - ∠ O P C = ∠ L P B - ∠ O P D$
$∴ ∠ C P A = ∠ D P B$

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In the figure, P is the centre of the circle. Two chords AB and CD are parallel to each other. Prove ∠CPA=∠DPB.

Consider ΔOPC and ΔOPDSince, OP=OP as a common side.∠POC=POD [Right angle]PC=PDThus, ΔPOC≅ΔPOD from RHS CriterionSo, ∠OPC=∠OPD.....(i)Similar way, we can prove ΔLPA≅ΔLPBSo, ∠LPA=∠LPB.....(ii)From equation(ii)-equation(i), we get ⇒∠LPA−∠OPC=∠LPB−∠OPD∴∠CPA=∠DPB

In the figure, $P$ is the centre of the circle. Two chords $A B$ and $C D$ are parallel to each other. Prove $∠ C P A = ∠ D P B$ .