Question

In Figure 3.11, line AB || line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of ∠BPQ and ∠PQD respectively. Prove that m ∠PTQ = 90°.

Answer 1

Your Answer is in the picture. Hope it he you man. Mark me as BRAINLIEST if you think.

Answer 2

Solution :

Given AB // CD , PQ is the

transversal.

Ray PT and Ray QT are angle

Bisectors of <BPQ and <PQD

respectively.

i ) <BPQ + <PQD = 180° ---( 1 )

[ Sum of the interior angles

same side of the transversal are

Supplementary ]

ii ) In ∆PTQ ,

<PTQ + <PQT + <TPQ = 180°

[ Angle sum property ]

<PTQ+ ( <PQD/2)+(<BQP/2)= 180°

=> <PTQ + ( <PQD + <BQP)/2 = 180°

=> <PTQ + 180°/2 = 180° [from(1)]

=> <PTQ + 90° = 180°

=> <PTQ = 180° - 90°

=> <PTQ = 90°

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