三、解答题(共52分)
9. (16分)如图,在$\mathrm{Rt}△ ABC$中,$∠ BAC=90°$,D是BC的中点,E是AD的中点,过点A作$AF// BC$交CE的延长线于点F.
(1)求证:四边形ADBF是菱形;
(2)若$AB=8$,菱形ADBF的面积为40,求AC的长.
答案：9.(1)证明:$\because AF// BC$,
$\therefore ∠ AFC=∠ FCD$,$∠ FAE=∠ CDE$.
$\because E$是$AD$的中点,$\therefore AE=DE$,
$\therefore △ FAE≌△ CDE(\mathrm{AAS})$,$\therefore AF=CD$.
$\because D$是$BC$的中点,$\therefore BD=CD$,$\therefore AF=BD$,
$\therefore$四边形$ADBF$是平行四边形.
$\because ∠ BAC=90°$,$D$是$BC$的中点,$\therefore AD=BD=\frac{1}{2}BC$,
$\therefore$四边形$ADBF$是菱形.
(2)解:$\because$四边形$ADBF$是菱形,
$\therefore$菱形$ADBF$的面积$=2△ ABD$的面积.
$\because D$是$BC$的中点,
$\therefore △ ABC$的面积$=2△ ABD$的面积,
$\therefore$菱形$ADBF$的面积$=△ ABC$的面积$=40$,
$\therefore \frac{1}{2}AB· AC=40$,$\therefore \frac{1}{2}×8· AC=40$,$\therefore AC=10$,
$\therefore AC$的长为10.
10. (18分)如图,在平行四边形ABCD中,E为线段CD的中点,连接AC,AE,延长AE,BC交于点F,连接DF,$∠ ACF=90°$.
(1)求证:四边形ACFD是矩形;
(2)若$CD=13$,$CF=5$,求四边形ABCE的面积.
答案：10.(1)证明:$\because$四边形$ABCD$是平行四边形,$\therefore AD// BC$,
$\therefore ∠ ADE=∠ FCE$,$∠ DAE=∠ CFE$.
$\because E$为线段$CD$的中点,$\therefore DE=CE$,
$\therefore △ ADE≌△ FCE(\mathrm{AAS})$,$\therefore AE=FE$,
$\therefore$四边形$ACFD$是平行四边形.
$\because ∠ ACF=90°$,$\therefore$四边形$ACFD$是矩形.
(2)解:$\because$四边形$ACFD$是矩形,
$\therefore ∠ CFD=90°$,$AC=DF$,$CF=AD$.
$\because$在$□ ABCD$中,$BC=AD$,$\therefore BC=CF=5$.
$\because CD=13$,$CF=5$,
$\therefore DF=\sqrt{CD^2-CF^2}=\sqrt{13^2-5^2}=12$.$\therefore AC=DF=12$.
$\because △ ADE≌△ FCE$,
$\therefore △ CEF$的面积$=△ ADE$的面积$=\frac{1}{2}△ ACF$的面
积$=\frac{1}{2}×\frac{1}{2}×5×12=15$,
平行四边形$ABCD$的面积$=BC· AC=5×12=60$,
$\therefore$四边形$ABCE$的面积$=$平行四边形$ABCD$的面积$-$
$△ ADE$的面积$=60-15=45$.
11. (18分)(2025·新疆)如图,在四边形ABCD中,$AD// BC$,BD是对角线.
(1)尺规作图:请用无刻度的直尺和圆规,作线段BD的垂直平分线,垂足为O,与边AD,BC分别交于点E,F;(要求:不写作法,保留作图痕迹)
(2)在(1)的条件下,连接BE,DF,求证:四边形BFDE为菱形.
答案：11.(1)解:如答图,直线$EF$即为所求.
(2)证明:$\because$直线$EF$是线段$BD$的垂直平分线,
$\therefore BE=DE$,$BF=DF$,$OB=OD$.
$\because AD// BC$,$\therefore ∠ EDO=∠ FBO$,$∠ DEO=∠ BFO$,
$\therefore △ ODE≌△ OBF(\mathrm{AAS})$,$\therefore DE=BF$,
$\therefore BE=DE=BF=DF$,$\therefore$四边形$BFDE$为菱形.
