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

Question:

In the below fig, AB ∥ CD and P is any point shown in the figure. Prove that: ∠ABP + ∠BPD + ∠CDP = 360°

Solution:

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

Now,

A8 || PM

⟹ ABP + BPM = 180

And

CD||PM = MPD + CDP = 180

Adding (i) and (ii), we get A8P + (BPM + MPD) CDP = 360

⟹ ABP + BPD + CDP = 360

Leave a comment

Close

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now