a)Ta có:\(\dfrac{AE}{AC}\)=\(\dfrac{2}{4}\)=\(\dfrac{1}{2}\)
\(\dfrac{AD}{AB}\)=\(\dfrac{3}{6}\)=\(\dfrac{1}{2}\)
nên:\(\dfrac{AE}{AC}\)=\(\dfrac{AD}{AB}\)
xét ΔADE và ΔACB có: \(\dfrac{AD}{AC}\)=\(\dfrac{AE}{AB}\)(CMT)
góc A chung
vậy ΔADE ∼ ΔACB(c.g.c)