In the figure, lines AB, CD and EF intersect at P. If ∠CPE = 56°, ∠APD=60° and PB = PF, then what is the measure of ∠PBF?

58°

We know that the sum of the angles of a straight line = 180^o

∠APD = ∠CPB ( vertically opposite angles)

∠CPB = 60°

∠CPB + ∠EPC + ∠BPF = 180°

60° + 56° + ∠BPF = 180°

∠BPF = 180° - 60° - 56° = 64°

Now, in ΔPBF

∠PBF + ∠PFB + ∠BPF = 180^o   and ∠PBF = ∠PFB

2∠PBF + 64° = 180°

∠PBF = 58^o