Help me for this math

(a) ∠BAE = ∠BAD + ∠DAE = 42 + 54 = 96
In ΔBAE, as AB = AE, so
∠AEB = (180 - 96)/2 = 42 ...... (1)In ΔDAE, as AD = AE, so∠AED = (180 - 54)/2 = 63 ...... (2)So, ∠BEC = ∠AEC - ∠AEB= 63 - 42 ...................... (from (1), (2))= 21
(b) In ΔAEC∠ACE= 180 - ∠EAC - ∠AEC= 180 - 96 - 63= 21 ...................,,,,,,,,,.......... (3)In ΔBCE, as ∠BCE = 21 (from (3)) and ∠BEC = 21 (from (a)), so, BE = BC, therefore ΔBCE is an isosceles triangle

2014-01-08 21:49:54 補充：
跟題目講，照著畫就是了。
先畫AB=AD=AE, BAD=42, DAE=54, 延長BA及ED, 交於C, 完成。

圖在哪？~~~~~~~~~~~~~

2014-01-08 17:42:45 補充：
那些年 :

大大，圖都沒有，你怎麽做的呀！