第43页 - 同步练习九年级数学苏科版江苏凤凰科学技术出版社

$解：(1)△ABE∽△ECD$

$∵EC//AB$

$∴∠A=∠CED$

$∵EB//DC$

$∴∠AEB=∠EDC$

$(2)∵△ABE∽△ECD$

$∴\frac {S_{△ABE}}{S_{△ECD}}=(\frac {h_{1}}{h_{2}})^2=\frac 49$

$∴\frac {h_{1}}{h_{2}}=\frac 23$

$∵\frac {S_{△BCE}}{S_{△CDE}}=\frac {BE}{CD}=\frac {h_{1}}{h_{2}}=\frac 23，S_{△CDE}=9$

$∴S_{△BCE}=6$

$解：图①中，设DE=x\ \mathrm {cm}，则DG=2x\ \mathrm {cm}$

$∵四边形DEFG 是矩形$

$∴DG//BC$

$∴△ADG∽△ABC$

$∴\frac {AM}{AH}=\frac {DG}{BC}$

$∵DG=2x\ \mathrm {cm}，BC=12\ \mathrm {cm}，AH=8\ \mathrm {cm}$

$∴\frac {AM}8=\frac {2x}{12}$

$∴AM=\frac 43x\ \mathrm {cm}$

$∵MH=DE=x\ \mathrm {cm}$

$又∵AM+MH=AH$

$∴\frac 43x+x=8$

$解得x=\frac {24}{7}$

$∴S_{矩形DEFG}=DE×DG=\frac {1152}{49}\ \mathrm {cm^2}$

$图②中，设DG=x\ \mathrm {cm}，则DE=2x\ \mathrm {cm}$

$同理可得，△ADG∽△ABC$

$∴\frac {AM}{AH}=\frac {DG}{BC}$

$∴AM=\frac 23x\ \mathrm {cm}$

$∵MH=DE=2x\ \mathrm {cm}，AM+MH=AH=8\ \mathrm {cm}$

$∴\frac 23x+2x=8$

$解得x=3$

$∴DG=3\ \mathrm {cm}，DE=6\ \mathrm {cm}$

$∴S_{矩形DEFG}=DG×DE=18\ \mathrm {cm^2}$

$∵\frac {1152}{49}＞18$

∴图①的设计方案更好

$解：由题意得AB=k_{AD}，A'B'=kA'D'$

$∵BD=\sqrt {AB^2-AD^2}=\sqrt {k^2-1}AD，B'D'=\sqrt {A'B'^2-A'D'^2}=\sqrt {k^2-1}A'D'$

$∴\frac {BD}{B'D'}=\frac {AD}{A'D'}$

$∵∠ADB=∠A'D'B'=90°$

$∴△ABD∽△A'B'D'$

$∴∠ABD=∠A'B'D'$

$∵∠C=∠C'$

$∴△ABC∽△A'B'C'$

$∴\frac {AB}{A'B'}=\frac {AD}{A'D'}=\frac {BE}{B'E'}$

$∴AD · B'E'=A'D' · BE$