PSLEBCDE is a parallelogram, BDF is an equilateral triangle and DE = EF.
- Find ∠BDE.
- Find ∠DEB.
- Find ∠FBC.
(a)
DE = EF
Triangle DEF is an isosceles triangle.
∠EDF
= (180° - 134°) ÷ 2
= 46° ÷ 2
= 23° (Isosceles triangle)
∠BDE
= 60° - 23°
= 37° (Equilateral triangle)
(b)
∠BFE
= 60° - 23°
= 37° (Equilateral triangle)
∠BEF
= 180° - 37° - 23°
= 120° (Angles sum of triangle)
∠BED
= 360° - 134° - 120°
= 106° (Angles at a point)
∠CBD
= ∠BDE
= 37° (Alternate angles)
∠CBF
= 60° + 37°
= 97°
Answer(s): (a) 37°; (b) 106°; (c) 97°