Question

in the figure angle PDQ =45 degree, angle PQD =35 degree and Angle BOP = 80 degree, prove that P//M​

Answer 1

Step-by-step explanation:

In △△PDQ  

⇒⇒∠∠PDQ + ∠∠PQD + ∠∠DPQ = 180          (Angle sum property)

⇒⇒45 + 35 + ∠∠DPQ = 180

⇒⇒80 + ​ ∠∠DPQ = 180

⇒⇒ ∠∠DPQ = 100             .................... (1)

Now,  

⇒⇒∠∠DPQ + ​∠∠QPM = 180                (Linear pair angles)

⇒⇒∠∠QPM = 180 - 100                       (From (1) )

⇒⇒∠∠QPM = 80            

Therefore, ​∠∠QPM = ​∠∠BOP = 80  

Hence, p is parallel to m since the corresponding angles are equal.

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Answer 2

$\huge\tt{\underline{\underline{Solution:}}}$

In ∆PDQ, we have

$\angle$DPQ + $\angle$PDQ + $\angle$PQD = 180°

$\implies$$\angle$DPQ + 45° + 35° = 180°.

$\implies$$\angle$DPQ+80° = 180°.

$\implies$$\angle$DPQ = 180°-80° = 100°.

Mark a point E on the line m such that QPE forms an angle. Since $\angle$QPE and $\angle$QPD forms a linear pair.

•°• $\angle$QPE + $\angle$QPD = 180°.

$\implies$$\angle$QPE+100°=180°.

$\implies$$\angle$QPE=180°-100°

$\implies$$\angle$QPE = 80°.

Now, p and m are two lines such that a transversal n intersect them O and P respectively such that the corresponding angles on the same side are equal, i.e.,

$\angle$AOB=$\angle$QPE=80°.

Hence , p ll m.