《2025年作业本江西教育出版社九年级数学全一册北师大版》

8. 如图所示，在$□ ABCD$中，已知$EF // AB$，$DE : AE = 2 : 3$.若$\bigtriangleup BDC$的面积为$50$，则四边形$AEFB$的面积为42.

答案： 42

9. 如图所示，在$\bigtriangleup ABC$中，$D$，$E$分别是边$AB$，$AC$上的点，已知$\angle BDE + \angle C = 180^{\circ}$. 求证：$\bigtriangleup ADE \backsim \bigtriangleup ACB$.

答案： 证明：$\because \angle BDE + \angle C = 180^{\circ}$，$\angle BDE + \angle ADE = 180^{\circ}$，
$\therefore \angle ADE = \angle C$.
又$\because \angle A = \angle A$，
$\therefore \triangle ADE \backsim \triangle ACB$.

10. 如图所示，在平行四边形$ABCD$中，$P$是线段$BC$的中点.连接$AC$交$DP$于点$E$，连接$BE$并延长，交$CD$于点$F$.
(1)求证：$F$为$CD$的中点.
(2)若$BF \perp CD$，且$BE = \sqrt{2}CE$，求$\frac{AB}{BC}$的值.

答案：
(1)证明：$\because$ 四边形$ABCD$是平行四边形，
$\therefore AB // CD$，$AD // BC$，$AB = CD$，$AD = BC$，
$\therefore \triangle ADE \backsim \triangle CPE$，$\triangle ABE \backsim \triangle CFE$，
$\therefore \frac{AE}{CE} = \frac{AD}{CP}$，$\frac{AB}{CF} = \frac{AE}{CE}$，$\therefore \frac{AB}{CF} = \frac{AD}{CP}$.
$\because P$是线段$BC$的中点，
$\therefore CP = \frac{1}{2}BC = \frac{1}{2}AD$，$\therefore \frac{AB}{CF} = \frac{AD}{CP} = 2$，
$\therefore \frac{CD}{CF} = \frac{AB}{CF} = 2$，$\therefore F$为$CD$的中点.
(2)解：由
(1)知$\triangle ABE \backsim \triangle CFE$，
$\therefore \frac{BE}{EF} = \frac{AB}{CF} = 2$.
设$EF = a$，则$BE = 2a$，
$\therefore BF = BE + EF = 3a$.
$\because BE = \sqrt{2}CE$，$\therefore CE = \sqrt{2}a$.
$\because BF \perp CD$，$\therefore$ 在$Rt\triangle EFC$中，由勾股定理，得$FC = \sqrt{EC^2 - EF^2} = a$，
$\therefore$ 在$Rt\triangle BFC$中，由勾股定理，得$BC = \sqrt{BF^2 + FC^2} = \sqrt{10}a$，$AB = 2CF = 2a$，
$\therefore \frac{AB}{BC} = \frac{2a}{\sqrt{10}a} = \frac{\sqrt{10}}{5}$.