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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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 ,
∠BPQ + ∠DQP =
⇒ ∠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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