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In the below fig, AB || CD and P is any point shown in the figure. Prove that:

`∠`ABP+`∠`BPD+`∠`CDP = 36O°

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#### Solution

Through P, draw a line PM parallel to AB or CD.

Now,

*AB *|| *PM ⇒* `∠`*ABP *+ `∠`*BPM *= 180°

And

*CD *|| *PM ⇒* `∠`*MPD *+ `∠`*CDP *= 180°

Adding (i) and (ii), we get

*`∠`ABP *+ (`∠`*BPM *+ `∠`*M**PD*) `∠`*CDP *= 360°

⇒`∠`*ABP *+ `∠`*BPD *+ `∠`*CDP *= 3 60°

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