Question

complete the following activity. In the given figure line AB II line CD line PQ intersect lines AB and CD in point R and S respectively
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Answer 1

Answer:

In the given figure,

∠PQD+∠QPB=180∘                         

Since PT and QT are bisectors of ∠BPQ and ∠PQD,

Hence,

21∠QPB=∠QPT, 21∠DQP=∠TQP

Dividing equation(1) in both side by 2

21×(∠PQD+∠QPB)=90∘

(∠PQT+∠QPT)=90∘             

Now,  in △PQT,

∠PQT+∠QPT+∠PTQ=180∘

90∘+∠PTQ=180∘                     [fromequation(2)]

∠PTQ=90∘