In the figure, ABCD is a trapezium and ADEH is a rhombus. AD is parallel to BC and DE = DF. HDC and DGF are straight lines. ∠DAH = 40° and ∠GDH = 35°. Find
- ∠ADC
- ∠DFE
(a)
∠ADH
= (180° - 40°) ÷ 2
= 70° (Isosceles triangle)
∠ADC
= 180° - 70°
= 110° (Angles on a straight line)
(b)
∠FDE
= 70° - 35°
= 35°
∠DFE
= (180° - 35°) ÷ 2
= 72.5 ° (Isosceles triangle)
Answer(s): (a) 110°; (b) 72.5°