I是△ABC的内心,过I作DE⊥AI,分别交AB,AC于点D,E,连接BI,CI,求证

证明：（1）∵DE⊥AI
∴∠CAI+∠AEI=90°
∵∠AEI=∠ICE+∠EIC
∴∠CAI+∠ICE+∠EIC=90°
∵I是△ABC的内心
∴∠CAI=1/2∠CAB∠ICE=1/2∠BCA∠IBC=1/2∠ABC
∵∠CAB+∠BCA+∠ABC=180°
∴1/2∠CAB+1/2∠BCA+1/2∠ABC=90°
∴∠CAI+∠ICE+∠IBC=90°
∴∠EIC=∠IBC
同理∠DIB=∠ICB
∵∠EIC=∠IBC∠ECI=∠ICB
∴⊿ECI∽⊿ICB
∴IC/CE=BC/CI
∴CI²=CE*BC
(2)∵∠EIC=∠IBC∠DIB=∠ICB
∵∠IBC=∠IBD∠ICB=∠ICE
∴∠EIC=∠DBI∠DIB=⊿ECI
∴⊿DIB∽⊿ECI
∴DI/DB=EC/EI
∴DI*EI=EC*BD
∵∠IAD=∠IAE∠AID=∠AIE=90°AI=AI
∴⊿AID≌⊿AIE
∴DI=EI
∴EI²=EC*BD