证明:过D作DE⊥AB,垂足为E,∴∠DEA=90°,∵∠ACB=90°,AC=BC,AD平分∠CAB.∴∠C=∠DEA,∠CAD=∠EAD,∴△ACD≌△AED;∴AC=AE,CD=DE;∵AC=BC,∠ACB=90°,∴∠B=45°,∴∠BDE=45°;∴CD=DE=EB;∴AB=AE+EB=AC+CD.
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