Answer :

Given: AB || CD, line PQ is the tranversal


Ray PT and Ray QT are bisectors of BPQ and PQD


To prove: PTQ = 90°


Proof: Since, Ray PT and Ray QT are bisectors of BPQ and PQD


TPQ = BPQ/2 ……..(1)


PQT = PQD/2 ………(2)


Since, two parallel lines are intersected by a transversal, the interior angles on either side of the transversal are supplementary.


So, BPQ + PQD = 180°


Dividing both sides by 2, we get


(BPQ + PQD)/2 = 180°/2


BPQ/2 + PQD/2 = 90°


In ΔPQT,


TPQ + PQT + PTQ = 180°


Substituting TPQ and PQT from (1) and (2) respectively


BPQ/2 + PQD/2 + PQT = 180°


90° + PQT = 180°


PQT = 180° - 90°


PQT = 90°


Hence, proved.


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