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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° - Mathematics

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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 + `∠`MPD) `∠`CDP = 360°

⇒`∠`ABP + `∠`BPD + `∠`CDP = 3 60°

Concept: Lines Parallel to the Same Line
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APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 10 Lines and Angles
Exercise 10.4 | Q 25 | Page 49

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