证明∵在△ABC中,AB=AC∴△ABC是等腰三角形，∴∠ABC=∠ACB∵AE=BE，且AE⊥BE∴ABE是等腰直角三角形∴∠BAE=∠ABE=45°∴∠ABC=∠ACB=67.5°∵AD平分∠BAC且平分BC∴∠BAD=∠CAD=∠ABC-∠ABE=22.5°∵∠AEB=∠BEC=90°∵AE=BE∴△AEH≌△BEC∴AH=BC∵D为中点∴AH=2BD
