Ta có: `\hatD_4 = \hatD = 110^o` (đối đỉnh)
Do $d'//d''$
`⇒ \hatE_1 = \hatC = 60^o` (so le trong)
`⇒ \hatG_2 = \hatD_1 = 110^o` (so le trong)
`⇒ \hatG_2 + \hatG_3 = 180^o` (kể bù)
Hay `110^o + \hatG_3 = 180^o`
`⇒ \hatG_3 = 70^o`
Do $d//d''$
`⇒ \hatE_1 = \hatA_5 = 60^o` (đồng vị)
`⇒ \hatG_3 = \hatB_6 = 70^o` (đồng vị)