Maths.....

∠DCA + ∠CAB = 180° ( int. ∠s, CD//AB )
CE is the angle bisector of ∠DCA,
therefore ∠ACE = ∠DCE
CE is the angle bisector of ∠CAB,
therefore ∠CAE = ∠EAB
∠DCA = ∠ACE + ∠DCE
∠CAB = ∠CAE + ∠EAB
( ∠ACE + ∠DCE ) + ( ∠CAE + ∠EAB ) = 180°
2( ∠ACE + ∠CAE ) = 180°
∠ACE + ∠CAE = 90°
∠AEC = 180° - ∠ACE - ∠CAE ( ∠ sum of ▲ )
∠AEC = 90°
Therefore CE is perpendicular to AE

2007-04-03 19:55:46 補充：
sorry~中間第四句打錯, 應為:AE is the angle bisector of ∠CAB,

2007-04-04 10:45:19 補充：
下面果位朋友, 你最後果句錯左喇你唔可以直接話 ∠AEC 等於 ∠ACE + ∠CAE 你都冇 reason!係要用 ∠ sum of ▲黎求架

let angleCEF=A ,angleFEA=B
let a point on CA be F such that FE//CD
therefore
angleCEF=angleDCF(alternate angle,CD//EF)
angleFEA=angleEAB(alternate angle,AB//EF)
therefore angle CEA=angleDCE+angleBAE=angleECA+angleEAC=A+B
therefore angle sum of triangle CEA=angleCEA+angleECA+angleCAE=2A+2B=180
therefore A+B=90
Since angle CEA=A+B,therefore angle CEA=90,therefore, CE is perpendicular to AE

2007-04-04 20:46:47 補充：
上面o個位朋友therefore angle sum of triangle CEA=angleCEA+angleECA+angleCAE=2A+2B=180呢句係由第6行到第7行到講左架,不要誤會

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because i can not upload the picture
so if your add me i will teach you by msn