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3.「2025安徽蚌埠模拟，★★☆」如图，在$△ ABC$中，点D在边BC上，连接AD并延长至点E，使$DE = \frac{5}{3}AD$，且$∠ AEB = ∠ C$，$AE = 16$，$BD=8$，则CD的长为（
A.$\frac{27}{4}$
B.$\frac{15}{2}$
C.$\frac{24}{5}$
D.$\frac{15}{8}$
B
）A.$\frac{27}{4}$
B.$\frac{15}{2}$
C.$\frac{24}{5}$
D.$\frac{15}{8}$
答案
B $\because DE=\frac{5}{3}AD,\therefore DE=\frac{5}{3+5}AE=10,AD=\frac{3}{3+5}AE=6$,
$\because ∠ AEB=∠ C,∠ BDE=∠ ADC,\therefore △ BDE∽ △ ADC$,
$\therefore \frac{CD}{DE}=\frac{AD}{BD},即\frac{CD}{10}=\frac{6}{8},\therefore CD=\frac{15}{2}$.故选 B.
$\because ∠ AEB=∠ C,∠ BDE=∠ ADC,\therefore △ BDE∽ △ ADC$,
$\therefore \frac{CD}{DE}=\frac{AD}{BD},即\frac{CD}{10}=\frac{6}{8},\therefore CD=\frac{15}{2}$.故选 B.
4.「2026浙江宁波期中，★★☆」如图,在矩形ABCD中,点E在AD上,AC与BE交于点F,若AB=3,AC=5,DE=3AE,求:
(1)AE的长.
(2)△ABF的面积.
(1)AE的长.
(2)△ABF的面积.
答案
(1)$\because$ 四边形 $ABCD$ 是矩形,$AB=3$,
$\therefore ∠ D=90°,CD=AB=3$,
$\therefore AD=\sqrt{AC^2-CD^2}=\sqrt{5^2-3^2}=4$,
$\because AE+DE=AD=4,且 DE=3AE$,
$\therefore AE+3AE=4,\therefore AE=1$.
(2)$\because$ 四边形 $ABCD$ 是矩形,$AD=4,\therefore AE// CB,BC=AD=4$,
$\therefore ∠ AEF=∠ CBF,∠ EAF=∠ BCF,\therefore △ AFE∽ △ CFB$,
$\therefore \frac{EF}{BF}=\frac{AE}{CB}=\frac{1}{4}$,
$\therefore BF=\frac{4}{1+4}BE=\frac{4}{5}BE$,
$\because ∠ BAE=90°,AB=3,AE=1$,
$\therefore S_{△ ABE}=\frac{1}{2}×3×1=\frac{3}{2}$,
$\therefore S_{△ ABF}=\frac{4}{5}S_{△ ABE}=\frac{4}{5}×\frac{3}{2}=\frac{6}{5}$.
$\therefore ∠ D=90°,CD=AB=3$,
$\therefore AD=\sqrt{AC^2-CD^2}=\sqrt{5^2-3^2}=4$,
$\because AE+DE=AD=4,且 DE=3AE$,
$\therefore AE+3AE=4,\therefore AE=1$.
(2)$\because$ 四边形 $ABCD$ 是矩形,$AD=4,\therefore AE// CB,BC=AD=4$,
$\therefore ∠ AEF=∠ CBF,∠ EAF=∠ BCF,\therefore △ AFE∽ △ CFB$,
$\therefore \frac{EF}{BF}=\frac{AE}{CB}=\frac{1}{4}$,
$\therefore BF=\frac{4}{1+4}BE=\frac{4}{5}BE$,
$\because ∠ BAE=90°,AB=3,AE=1$,
$\therefore S_{△ ABE}=\frac{1}{2}×3×1=\frac{3}{2}$,
$\therefore S_{△ ABF}=\frac{4}{5}S_{△ ABE}=\frac{4}{5}×\frac{3}{2}=\frac{6}{5}$.
