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