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.

#### Solution

Given that *AB *|| *CD*

Let *E**F *be the parallel line to AB and CD which passes through P.

It can be seen from the figure

Alternative angles are equal

`∠`*AB**P *= `∠`*BP**F*

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°