Question

In Fig 8.113, lines AB and CD are parallel and P is any point as shown in the figure. Show that ∠ABP +∠CDP = ∠DPB.

Answer 1

Proved below.

Step-by-step explanation:

Given:

Here AB║CD and P is any point as shown in the figure.

Construction:

Draw line XY passing through point P and parallel to AB and CD.

To prove:

∠ABP +∠CDP = ∠DPB.

Proof:

As shown in the figure,

XY║CD

∴ ∠DPY = ∠CDP      [alternate interior opposite angles]   [1]

Also XY║AB

∴ ∠BPY = ∠ABP      [alternate interior opposite angles]   [2]

Now adding Eq (1) and (2), we get

∠DPY +∠BPY = ∠CDP + ∠ABP

∠DPB = ∠CDP + ∠ABP     [shown in the figure]

Hence proved.

Answer 2

Answer:

Given:

Here AB║CD and P is any point as shown in the figure.

Construction:

Draw line XY passing through point P and parallel to AB and CD.

To prove:

∠ABP +∠CDP = ∠DPB.

Proof:

As shown in the figure,

XY║CD

∴ ∠DPY = ∠CDP      [alternate interior opposite angles]   [1]

Also XY║AB

∴ ∠BPY = ∠ABP      [alternate interior opposite angles]   [2]

Now adding Eq (1) and (2), we get

∠DPY +∠BPY = ∠CDP + ∠ABP

∠DPB = ∠CDP + ∠ABP    

Hope it helps