Question

In the below fig, lines AB and CD are parallel and P is any point as shown in the figure.
Show that` ∠`ABP +` ∠`CDP = ∠DPB.

Answer 1

Given that AB || CD

Let EF be the parallel line to AB and CD which passes through P.

It can be seen from the figure

Alternative angles are equal

`∠`ABP = `∠`BPF

Alternative angles are equal

`∠`CDP = `∠`DPF

⇒ `∠`ABP + `∠`CDP = `∠`BPF + `∠`DPF

⇒ `∠`ABP + `∠`CDP = `∠`DPB

Hence proved

AB parallel to CD, P is any point

To prove: `∠`ABP + `∠`BPD + `∠`CDP = 360°

Construction : Draw  EF || AB passing through P

Proof:

Since AB || EF and  AB || CD

∴EF || CD       [Lines parallel to the same line are parallel to each other]

 `∠`ABP + `∠`EPB =180°      [Sum of co-interior angles is180° AB || EF and BP is the transversal]

 `∠`EPD + `∠`COP =180°

[Sum of co-interior angles is180° EF || CD and DP is transversal]                   …....(1)

 `∠`EPD + `∠`CDP =180°

[Sum of Co-interior angles is 180° EF || CD and DP is the transversal]                        …(2)   

By adding (1) and (2)

 `∠`ABP + `∠`EPB + `∠`EPD + `∠`CDP = 180° +180°

 `∠`ABP + `∠`EPB + `∠`COP = 360°