Bài 3:
a: Xét ΔABC có DE//BC
nên \(\dfrac{AD}{AB}=\dfrac{AE}{AC}\)
Xét ΔABC có DE//BC
nên \(\dfrac{AD}{DB}=\dfrac{AE}{EC}\)
=>\(\dfrac{BD}{AD}=\dfrac{EC}{AE}\)
b: Ta có: AD+DB=AB
=>AD+2=3
=>AD=1(cm)
Xét ΔABC có DE//BC
nên \(\dfrac{AD}{DB}=\dfrac{AE}{EC}\)
=>\(\dfrac{AE}{EC}=\dfrac{1}{2}\)
=>\(\dfrac{EC}{AE}=\dfrac{2}{1}\)
=>\(\dfrac{EC+AE}{AE}=\dfrac{2+1}{1}\)
=>\(\dfrac{AC}{AE}=\dfrac{3}{1}\)
=>\(\dfrac{AE}{AC}=\dfrac{1}{3}\)
mà AE+AC=10
nên \(AE=\dfrac{1}{4}\cdot10=2,5\left(cm\right);AC=10-2,5=7,5\left(cm\right)\)
EC+AE=AC
=>EC+2,5=7,5
=>EC=5(cm)
Bài 1:
a: AE+CE=16
mà AE+CE=AC
nên AC=16(cm)
AE+CE=16
=>AE+12=16
=>AE=4(cm)
\(\dfrac{AD}{AB}=\dfrac{3}{12}=\dfrac{1}{4};\dfrac{AE}{AC}=\dfrac{1}{4}\)
b: ta có: \(\dfrac{AD}{AB}=\dfrac{1}{4}\)
\(\dfrac{AE}{AC}=\dfrac{1}{4}\)
Do đó: \(\dfrac{AD}{AB}=\dfrac{AE}{AC}\)
