Math, asked by jnandini512, 14 days ago

In figure 3.11 ,line AB are parallel to look ne CD and line PQ is the transversal .Ray PT and Ray QT are bisectors of angle BPQ and Angle PQD respectively .

Prove that m angle PTQ =90

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Answers

Answered by anuj05slg
1

Answer:

AB || CD and PQ is a transversal line.

∠BPQ + ∠DQP = 180∘ (Angles on the same side of a transversal line are

                                       supplementary angles)

Now, dividing both sides by  \frac{1}{2},

\frac{1}{2}∠BPQ +   \frac{1}{2}∠DQP = \frac{180}{2}

⇒ ∠QPT + ∠PQT = 90∘

In ΔPQT,

∠QPT + ∠PQT + ∠PTQ = 180° (sum of all ∠s of a triangle is 180°)

90° + ∠PTQ = 180°

∠PTQ           = 180 - 90

∠PTQ           = 90     [ Proved]

Step-by-step explanation:

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