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?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 58°
Two facts decide it: vertically opposite angles are equal, and in a triangle with two equal sides the angles opposite those sides are equal.
\(\angle BPC=\angle APD=60^\circ\) (vertically opposite angles)
E, P and F lie on one line, so \(\angle CPE+\angle BPC+\angle BPF=180^\circ\)
\(\angle BPF=180^\circ-56^\circ-60^\circ=64^\circ\)
PB = PF, so in triangle PBF, \(\angle PBF=\angle PFB\)
\(2\angle PBF=180^\circ-64^\circ=116^\circ\), so \(\angle PBF=\) 58°
