In the figure, LDEH is a trapezium and triangles LHG and GLF are isosceles triangles. KC, KE and CE are straight lines. LH = LG = GF. Find

1. ∠i
2. ∠j

(a)
∠DLF = 180° - ∠i (Interior angles, LD//KE)
∠GFL = 180° - ∠i (Angles on a straight line)
∠GLF = 180° - ∠i (Isosceles triangle)

58° + 180° - ∠i + 180° - ∠i + ∠j + 16° = 180° (Angles on a straight line, CK)
58° + 180° + 180° + 16° - ∠i - ∠i + ∠j  = 180°
434° - 2∠i + ∠j = 180°
2∠i - ∠j = 434° - 180° 
2∠i - ∠j = 254°

∠j = 2∠i - 254° --- (1)

∠LGH
= ∠LHG
= 2 x (180° - ∠i)
= 360° - 2∠i (Exterior angle of a triangle)

∠j = 180° - (360° - 2∠i) - (360° - 2∠i)
∠j = 180° - 360° + 2∠i - 360° + 2∠i 
∠j = 180° - 360° - 360° + 2∠i + 2∠i
∠j = 4∠i - 540° (Angles sum of triangle) 

∠j = 4∠i - 540° --- (2)

(2) = (1)
4∠i - 540° = 2∠i - 254°
4∠i - 2∠i= 540° - 254°
2∠i = 286°

∠i
= 286° ÷ 2
= 143°

(b)
From (1)
∠j
=  2∠i - 254°
= 286° - 254° 
= 32°

Answer(s): (a) 143°; (b) 32°