证明：过点D作DE⊥AB于E,DF⊥AC于F∵∠BAC＝90,AB＝AC∴∠B＝∠C＝45∵DE⊥AB∴DE＝BE＝BD×√2/2∵DF⊥AC∴DF＝CF＝CD×√2/2∵∠BAC＝90,DE⊥AB,DF⊥AC∴矩形AEDF∴AE＝DF∵AD²＝AE²+DE²∴AD²...