Question

In the below fig, AB || CD and P is any point shown in the figure. Prove that:
`∠`ABP+`∠`BPD+`∠`CDP = 36O°

Answer 1

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°