Question

In the given figure AB||CD.prove that angle BPO+angle POQ+angle DQO=360°

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

From the figure ,

AB // CD

Construction:

Draw EF // AB //CD .

$\underline { To \:proof }$

$\angle BPO + \angle POQ + \angle DQO = 360\degree$

Proof:

i) AB//EF, PO is a transversal.

$\angle BPO + \angle POF = 180\degree$

$\blue { ( Sum \: of \:the \: interior\: angles \:of \:same}$

$\blue { side \:of \: the \: transversal \:are \: supplementary)}$

ii) EF//CD , OQ is a transversal.

$\angle FOQ + \angle OQD = 180\degree \:--(2)$

$Add \: (1) \: and \: (2) , \:we \:get$

$\angle BPO + \angle POF + \angle FOQ + \angle OQD = 180\degree + 180\degree$

$\angle BPO +( \angle POF + \angle FOQ) + \angle OQD = 360\degree$

$\angle BPO + \angle POQ + \angle DQO = 360\degree$

Hence , proved .

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