In Fig 8.114, AB||CD and P is any point shown in the figure. Prove that:

∠ABP + ∠BPD + ∠CDP = 360°

AB is parallel to CD, P is any point

To prove: ∠ABP + ∠BPD + ∠CDP = 360^o

Construction: Draw EF ‖ AB passing through F

Proof: Since,

AB ‖ EF and AB ‖ CD

Therefore,

EF ‖ CD (Lines parallel to the same line are parallel to each other)

∠ABP + ∠EPB = 180^o(Sum of co interior angles is 180^o, AB ‖ EF and BP is transversal)

∠EPD + ∠COP = 180^o (Sum of co. interior angles is 180^o, EF ‖ CD and DP is transversal) (i)

∠EPD + ∠CDP = 180^o(Sum of co. interior angles is 180^o, EF ‖ CD and DP is the transversal) (ii)

Adding (i) and (ii), we get

∠ABP + ∠EPB + ∠EPD + ∠CDP = 360^o

∠ABP + ∠EPD + ∠COP = 360^o