1.证明：∵∠ACB=90°∴AC⊥BC∵BF⊥CE∴∠ACE=∠CBG∵∠AEC=∠ADC+∠DCE=90°+∠DCE,∠BGC=∠GFC+∠DCE=90°+∠DCE∴∠AEC=∠BGC∵AC=BC∴△ACE≌△CBG∴AE=CG2.BE=CM证明：∵BF⊥CH,AC⊥BC∴∠ACH=∠CBF∵AC=BC∴R...