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Question

In the figure, $$P$$ is the centre of the circle. Two chords $$AB$$ and $$CD$$ are parallel to each other. Prove $$\angle CPA=\angle DPB$$.
1296661_4f3ef765fed742649d53578f5bf70887.png


Solution

$$\begin{array}{l} CD\parallel AB\, \, \left( { given } \right)  \\ \Rightarrow Cp=OP\, \, \left( { radius } \right)  \\ \Rightarrow AP=PB\, \, \, \left( { radius } \right)  \\ \Rightarrow \therefore \angle PCD=\angle PAB \\ \Rightarrow PDC=\angle PBA \\ hence,\, =\angle DPB \end{array}$$

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