依题意△ADE为等腰直角三角形故∠AEC=∠ADB=25°(1) ∵AD=AE，∠EAD=90°∴△AEC≌△ABD∴AB=AC(2) ∵AB=AC，AD=AE ∠BAE=∠BAC+∠BAD=90°+∠BAD ∠CAD=∠EAD+∠BAD=90°+∠BAD ∴∠BAE=∠CAD ∴△ABE≌△ACD 故 BE=CD
