PC bisects BPD find M Is this right?

Question

asked Oct 11, 2019 32.2k views

2 Answers

Tahir Raza · Answer 1 · 2019-10-15T10:57:56+0000

Solving the problem in the following way

$\overrightarrow{PC}$ bisects ∠BPD.

Then, ∠BPC=∠DPC-----(1)

AD is a line , on which ray PB and PC lies.

→∠APB+∠BPC+∠CPD=180°----By linear pair axiom

→124°+∠DPC+∠DPC=180°------[Using 1]

→124°+2∠DPC=180°

→∠DPC=180°-124°

→2∠DPC=56°

Dividing both sides by,2 we get

→∠DPC=28°