第35页 - 苏科版九年级数学课课练答案（上下册）

$解：(1)\frac {AB}{BC}=\frac {3}{4}，\frac {CF}{CD}=\frac {4-\frac {7}{4}}{3}=\frac {3}{4}$

$(2)因为\frac {AB}{BC}=\frac {CF}{CD}=\frac {3}{4}$

$所以线段AB，BC，CF，CD成比例$

$解：(1)DB=12-AD$

$所以\frac {AD}{12-AD}=\frac {AE}{EC}=\frac {6}{4}$

$所以AD=\frac {36}{5}$

$(2)DB=12-\frac {36}{5}=\frac {24}{5}，AC=6+4=10$

$所以\frac {DB}{AB}=\frac {\frac {24}{5}}{12}=\frac {2}{5}$

$\frac {EC}{AC}=\frac {4}{10}=\frac {2}{5}$

$所以\frac {DB}{AB}=\frac {EC}{AC}$

$解：设S_{△COD}= x ， S_{△AOE}=y，$

$则\frac {35+x}{30}=\frac {OC}{OF}=\frac {y+84}{40}$

$即4x- 3y= 112$

$\frac {x}{y+84}=\frac {OD}{OA}=\frac {35}{30+40}$

$即2x-y= 84$

$解得x= 70，y= 56$

$所以S_{△ABC}= 40+ 30+ 35+ 70+84+ 56= 315$