Question

AB is parallel to CD . the bisector of angle BPQ and angle DQP meet at O. prove that angle POQ =90 degree

Answer 1

Given: AB is parallel to CD. The bisector of ∠BPQ and ∠ DQP meet at O.

To Prove: ∠POQ =90°

Proof: Since PO is bisector of ∠BPQ

∠BPO = ∠OPQ = (1/2) ∠BPQ -------------(i)

Again, QO is bisector of ∠DQP

∠DQO = ∠OQP = (1/2) ∠DQP-------------(ii)

Also, ∠BPQ + ∠DQP = 180°   ----------[Sum of interior angles between // lines]

So,  (1/2)∠BPQ + (1/2)∠DQP =(1/2) 180°

Now, by using equation (i) and (ii) in above, we get,

or,             ∠OPQ + ∠OQP = 90°   ------- (iii)

Now, we shall consider ΔOPQ,

           ∠OPQ + ∠OQP + ∠POQ = 180°    ---------sum of angle property

or,                  90° + ∠POQ = 180°     -  --------using equation (iii)

or,                           ∠POQ = 180° - 90° = 90°

Hence,                   ∠POQ =  90°

Answer 2

Answer:

hi

Step-by-step explanation:

Given: AB is parallel to CD. The bisector of ∠BPQ and ∠ DQP meet at O.

To Prove: ∠POQ =90°

Proof: Since PO is bisector of ∠BPQ

∠BPO = ∠OPQ = (1/2) ∠BPQ -------------(i)

Again, QO is bisector of ∠DQP

∠DQO = ∠OQP = (1/2) ∠DQP-------------(ii)

Also, ∠BPQ + ∠DQP = 180°   ----------[Sum of interior angles between // lines]

So,  (1/2)∠BPQ + (1/2)∠DQP =(1/2) 180°

Now, by using equation (i) and (ii) in above, we get,

or,             ∠OPQ + ∠OQP = 90°   ------- (iii)

Now, we shall consider ΔOPQ,

          ∠OPQ + ∠OQP + ∠POQ = 180°    ---------sum of angle property

or,                  90° + ∠POQ = 180°     -  --------using equation (iii)

or,                           ∠POQ = 180° - 90° = 90°

Hence,                   ∠POQ =  90°