Question

In Fig 8.114, AB||CD and P is any point shown in the figure. Prove that:
∠ABP + ∠BPD + ∠CDP = 360°

Answer 1

Given : AB || CD and P is any point shown in the figure.

 

To Prove : ∠ABP + ∠BPD + ∠CDP = 360°

Proof:  

Construction: Through P draw a line PM ‖ AB or  CD.

Since,  AB ‖  PM and AB ‖ CD

Therefore,

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

Since, AB ‖ PM and BP is a transversal and sum of the interior angles on the same side of a transversal is 180°.

∠ABP + ∠BPM = 1180°  ……………(1)

Since, CD ‖ PM and DP is a transversal and sum of the interior angles on the same side of a transversal is 180°.

∠MPD + ∠CDP = 180° …………….(2)

Adding eq (1) and (2), we get

∠ABP + (∠BPM  + ∠MPD )+ ∠CDP  = 360°  

∠ABP + ∠BPD + ∠CDP = 360°  

Hence proved.

HOPE THIS ANSWER WILL HELP YOU…..

 

Some questions of this chapter :

In Fig 8.112, if l||m, n||p and ∠1 =85°, find ∠2.

https://brainly.in/question/15905987

If, l, m, n are three lines such that l||m and n⊥ l, prove that n⊥ m.

https://brainly.in/question/15905997

Answer 2

Step-by-step explanation:

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TO PROVE :

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