第2个回答 2013-02-16
(1)证明:
∵∠CAB+∠B=90°,CF⊥AB
∴∠CAB+∠ACF=90°
∴∠ACF=∠B
∵AD⊥DE
∴∠BDE+∠ADC=90°
∴∠CAD=∠BDE
∵∠CAD=∠BDE,∠B=∠ACF
∴△ACG∽△DBE
(2)解:
延长AD,做BH⊥AD,垂足为H
∵ED⊥AD,BH⊥AD
∴DE/AD=BH/AH
∵CD=BD=BC/2,AC=BC/2
∴AC=CD=BD
∵∠ACB=90°
∴∠BDH=∠ADC=45°,AD=CD*√2
∵BH⊥AH
∴BH=DH=BC*√2/2
∵CD=BD
∴AD/BH=2,即AD=2*BH
∵BH=DH
∴BH/AH=BH/(DH+AH)=BH/(BH+2*BH)=1/3