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

1.	In Fig, AB\|\|CD and P is any point shown in the figure. Prove that:∠ABP + ∠BPD + ∠CDP = 360°
Answer» AB is parallel to CD, P is any point To prove: ∠ABP + ∠BPD + ∠CDP = 360° 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°(Sum of co interior angles is 180°, AB ‖ EF and BP is transversal) ∠EPD + ∠COP = 180°(Sum of co. interior angles is 180°, EF ‖ CD and DP is transversal) (i) ∠EPD + ∠CDP = 180°(Sum of co. interior angles is 180°, EF ‖ CD and DP is the transversal) (ii) Adding (i) and (ii), we get ∠ABP + ∠EPB + ∠EPD + ∠CDP = 360° ∠ABP + ∠EPD + ∠COP = 360°

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

Answer»

AB is parallel to CD, P is any point

To prove:

∠ABP + ∠BPD + ∠CDP = 360°

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°(Sum of co interior angles is 180°, AB ‖ EF and BP is transversal)

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

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

Adding (i) and (ii), we get

∠ABP + ∠EPB + ∠EPD + ∠CDP = 360°

∠ABP + ∠EPD + ∠COP = 360°