Firstly, we will draw figure
Let's assume
length of AC=x
we have
AB : BC = 2 : 1
so,
[tex]AB=\frac{2}{3} x[/tex]
[tex]BC=\frac{1}{3} x[/tex]
Point D is on line AB
and
[tex]AB=\frac{2}{3} x[/tex]
AD : DB = 3 : 2
so, we get
[tex]AD=\frac{3}{5} *\frac{2}{3} x[/tex]
[tex]AD=\frac{2}{5} x[/tex]
[tex]DB=\frac{2}{5} *\frac{2}{3} x[/tex]
[tex]DB=\frac{4}{15} x[/tex]
now, we can locate these values
Firstly, we will find DC
DC=DB+BC
now, we can plug values
[tex]DC=\frac{4}{15} x+\frac{1}{3} x[/tex]
[tex]DC=\frac{3x}{5}[/tex]
we have got
[tex]AD=\frac{2}{5} x[/tex]
now, we can find ratio
[tex]\frac{AD}{DC} =\frac{\frac{2}{5} x}{\frac{3x}{5}}[/tex]
now, we can simplify it
[tex]\frac{AD}{DC} =\frac{2}{3}[/tex]
so,
AD:DC=2:3...........Answer
