Bài 3:
Xét ΔABC có DE//BC
nên \(\dfrac{DE}{BC}=\dfrac{AD}{AB}\)
=>\(\dfrac{DE}{7,5}=\dfrac{2}{5}\)
=>\(DE=7,5\cdot\dfrac{2}{5}=\dfrac{15}{5}=3\left(cm\right)\)
Bài 2:
Xét ΔBAC có ED//AC
nên \(\dfrac{BE}{EA}=\dfrac{BD}{DC}\)
=>\(\dfrac{BE+EA}{EA}=\dfrac{BD+CD}{CD}\)
=>\(\dfrac{BA}{EA}=\dfrac{BC}{CD}\)
=>\(\dfrac{AE}{AB}=\dfrac{DC}{CB}\)
Xét ΔCAB có DF//AB
nên \(\dfrac{CF}{FA}=\dfrac{CD}{DB}\)
=>\(\dfrac{CF+FA}{FA}=\dfrac{CD+DB}{DB}\)
=>\(\dfrac{AC}{AF}=\dfrac{BC}{BD}\)
=>\(\dfrac{AF}{AC}=\dfrac{BD}{BC}\)
\(\dfrac{AE}{AB}+\dfrac{AF}{AC}=\dfrac{BD}{BC}+\dfrac{CD}{BC}=\dfrac{BC}{BC}=1\)
Bài 1:
Xét ΔABC có DE//BC
nên \(\dfrac{AD}{DB}=\dfrac{AE}{EC}\)
=>\(\dfrac{AE}{EC}=\dfrac{5}{3}\)
=>\(AE=\dfrac{5}{3}EC\)
mà AE-EC=3
nên \(\dfrac{5}{3}EC-EC=3\)
=>\(\dfrac{2}{3}EC=3\)
=>\(EC=3:\dfrac{2}{3}=3\cdot\dfrac{3}{2}=4,5\left(cm\right)\)
=>AE=4,5+3=7,5cm
AE+EC=AC
=>AC=7,5+4,5
=>AC=12(cm)