证
在△EBA和△EDC中
∠DEC+∠CEB+∠AEB=180°
∵△ABC为等腰三角形
∴∠ABC+∠BCA+∠CAB=180°
∴∠BCA=∠CAB
∵∠CEB=∠CAB (同弧圆周角)
∵∠AEB=∠ACB (同弧圆周角)
∴∠AEB=∠DEC (等量代换)
∠EDC+∠EAB+∠ABC=180°
∴∠ABC+∠ACB+∠CAB=180°
∴∠ABC+∠ABC+∠CAB=180°
∵∠EBC=∠CAE (同弧圆周角)
∴(∠ABC-∠EBC) +(∠CAB +∠CAE )+∠ABC=180°
∴∠EBA +∠EAB +∠ABC=180°
∴∠EDC=∠EBA
∴△EBA∽△EDC
∴AB/CD =AE/CE
∴AB×CE=AE×CD
证毕