WAEC - Mathematics (1995 - No. 35)

In the diagram above, AB//CD, the bisector of ∠BAC and ∠ACD meet at E. Find the value of ∠AEC

30^o

45^o

60^o

75^o

90^o

Ae is a bisector of BAC => EAC = 45^o
CE is the bisector of ACD => ACE = 45^o
AEC = 180^o - [EAC + ACE]
= 180^o - (45^o + 45^o) = 180^o - 90^o = 90^o