Question

已知如图,梯形abcd中,dc//ab,对角线ac=ab=9,dc=5,ad=6,过点D作BC的平行线交AC于点

Answer 1

（1）∵ CD∥AB DF∥BC∴ BF=CD=5∵ AB=AC∴ ∠B=∠ACB=∠CED=∠CDECE=BF=5AE=AC-CE=AB-CD=9-5=4AC∶AD=9∶6=3∶2AE∶AD=6∶4=3∶2∵∠CAD是公共角∴ △ADE∽△ACD（2）∵ △ADE∽△ACD∴ AD/AC=DE/CDDE=AD*CD/AC=6*5/9=10/3∵ ∠ACD=∠BAC ∠CDF=∠B∴ △CDE∽△ABCCD/AB=DE/BCBC=AB*DE/CD= 9*(10/3) / 5 = 6