In the figure, BCD is parallel to EFG and ED cuts ∠CEF equally. Given that ∠BCE = 56° and ∠CHD = 109°, find
- ∠a
- ∠b
(a)
∠BCE = ∠CEF (Alternate angles)
∠a
= 56° ÷ 2
= 28°
(b)
∠CEH = ∠a = 28°
∠b
= 109° - 28°
= 81° (Exterior angle of a triangle)
Answer(s): (a) 28°; (b) 81°