In Fig. 16.139, i

We have,

∠ACB = 40^o

∠DPB = 120^o

∠ADB = ∠ACB = 40^o (Angle on same segment)

In triangle PDB, by angle sum property

∠PDB + ∠PBD + ∠BPD = 180^o

40^o + ∠PBD + 120^o = 180^o

∠PBD = 20^o

Therefore,

∠CBD = 20^o