Geometry Practice Test 5

AB = AE and ∠CDE = 15°
⇒ ∠ABC = 65° = ∠AEB
AB || CD
BC is a transversal line
∠ABC = ∠BCD = 65°
In ΔCED,
∠CED+∠EDC+∠DCE = 180°
∠CED = 180° − (65° +15°)
∠CED = 100°
∠CED = ∠BEF (Vertically opposite angles)
∠CED = ∠AEB+∠AEF
65° + ∠AEF = 100°
∠AEF = 100° − 65° = 35°