证明:过点D作DE⊥AB于E
∵AD平分∠BAC,AC⊥CD,DE⊥AB
∴AE=AC (角平分线性质),∠AED=∠BED=90
∵AD=BD,DE=DE
∴△AED≌△BED (HL)
∴BE=AE
∴AB=AE+BE=2AE
∴AB=2AC