设EF=5x，则HE=9x，∵矩形EFGH内接于△ABC且AD⊥BC∴EH∥BC，EF∥AD∴△AEH∽△ABC，△BFE∽△BDA∴HEBC＝AEAB，EFAD＝BEAB，即9x48＝AEAB，5x16＝BEAB，∴9x48+5x16=AEAB+BEAB=ABAB=1．解得：x=2，∴矩形的周长为：...