Q. 8

# In Figure 3.11, l

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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