Question

AB IS PARALLEL TO CD.THE BISECTOR OF ANGLE BPQ AND ANGLE DQP MEET AT O.PROVE THAT ANGLE POQ=90DEGREE​

Answer 1

SoluTion :-

PO is bisector of ∠BPQ

∴ ∠BPO = ∠OPQ = Half of ∠BPQ___ (Eq. 1)

Also,

QO is bisector of ∠DQP

∴ ∠DQO = ∠OQP = (1/2) ∠DQP___ (Eq. 2)

Also,

∠BPQ + ∠DQP = 180° (Sum of interior angles between parallel lines)

∴ Half of ∠BPQ + Half of ∠DQP = Half of 180°

Now,

∠OPQ + ∠OQP = 90°___ (Eq. 3)

Also,

∠OPQ + ∠OQP + ∠POQ = 180° (Sum of angle property)

Thus,

90° + ∠POQ = 180° (Using Eq. 3)

∠POQ = 180° - 90° = 90°

∠POQ =  90°

Hence Proved