解：连接BE、BC、OC，BE交AC于F交OC于H．

∵AB是直径，∴∠AEB =90°，∴∠AEB=∠DEH=∠D=∠DCH=90°，∴四边形DEHC是矩形，∴∠EHC=90°，即OC⊥EB，∴DC=EH=HB，DE=HC，∵cos∠CAD= $\frac{4}{5}$ = $\frac{A D}{A C}$ ，设AD=4a，AC=5a，则DC=EH=HB=3a，∵cos∠CAB= $\frac{4}{5}$ = $\frac{A C}{A B}$ ，∴AB= $\frac{25}{4}$ a，BC= $\frac{15}{4}$ a，在Rt△CHB中，CH= $\sqrt{C B^{2} - B H^{2}}$ = $\frac{9}{4}$ a，∴DE=CH= $\frac{9}{4}$ a，AE= $\sqrt{A B^{2} - B E^{2}}$ = $\frac{7}{4}$ a，∵EF∥CD，∴ $\frac{A F}{F C}$ = $\frac{A E}{E D}$ = $\frac{7}{9}$ ．

福建省龙岩市永定县金丰片2018届九年级数学中考一模试卷