第52页 - 苏科版九年级（初三）数学学习与评价答案（上下册）

$证明：∵△ABC∽△A'B'C'$

$又AD、BE是△ABC的高，A'D'、B'E'是△A'B'C'的高$

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

$∴\frac {AD}{A'D'}=\frac {BE}{B'E'}$

$解：连接AP并延长交BC于点D$

$∵点P为△ABC的重心$

$∴\frac {AP}{AD}=\frac 23$

$∵EF//BC$

$∴△AEP∽△ABD，△AEF∽△ABC$

$∴\frac {AE}{AB}=\frac {AP}{AD}=\frac 23$

$∴\frac {EF}{BC}=\frac {AE}{AB}=\frac 23$

$解：∵DG//AB、QH//BC $

$∴△PKQ∽△DPE$

$∴\frac {S_{△KQP}}{S_{△PDE}}=(\frac {KP}{PE})^2=\frac 4{16}$

$∴\frac {KP}{PE}=\frac 12 $

$∴\frac {KP}{KE}=\frac 13$

$又∵△KQP∽△KBE$

$∴\frac {S_{△KQP}}{S_{△KBE}}=(\frac {KP}{KE})^2=(\frac 13)^2=\frac 19$

$∴\frac {4}{S_{△KBE}}=\frac 19 $

$∴S_{△KBE}=36$

$∴S_{四边形BDPQ}=S_{△KBE}-S_{△KQP}-S_{△PDE}=36-4-16=16$

$同理可求得S_{四边形CEPH}=24，S_{四边形AKPG}=12$

$∴S_{△ABC}=16+12+24+16+9+4=81$

$证明：如图四边形ABCD∽四边形A'B'C'D'$

$设相似比为k，则\frac {AD}{A'D'}=\frac {DC}{D'C'}=k，∠D=∠D'$

$∴△ADC∽△A'D'C'$

$∴\frac {AC}{A'C'}=\frac {AD}{A'D'}=k$

命题得证

解：它们对应顶点相连形成的直线交于一点