In the figure, BCDE is a trapezium and BEFJ is a rhombus. BE is parallel to CD and EF = EG. JED and EHG are straight lines. ∠EBJ = 50° and ∠HEJ = 35°. Find
- ∠BED
- ∠EGF
(a)
∠BEJ
= (180° - 50°) ÷ 2
= 65° (Isosceles triangle)
∠BED
= 180° - 65°
= 115° (Angles on a straight line)
(b)
∠GEF
= 65° - 35°
= 30°
∠EGF
= (180° - 30°) ÷ 2
= 75 ° (Isosceles triangle)
Answer(s): (a) 115°; (b) 75°