Auxiliary similar triangles [Thales' theorem and the theorem of proportional segments]

In triangle $ABC$, the bisector $AD$ divides side $BC$ in the ratio $BD: DC=2: 1$. In what ratio does the median from $C$ divide this bisector?
Let the median $C E$ and the bisector $A D$ intersect at point $M$.

First method. Draw a line through vertex $A$ parallel to $B C$ and extend the median $C E$ until it intersects this line at point $T$. From the equality of triangles $A E T$ and $B E C$, it follows that $A T = B C = 3 C D$, and from the similarity of triangles $A M T$ and $D M C$, we have $A M: M D = A T: C D = 3: 1$.

Second method. Draw a line through point $D$ parallel to $C E$ until it intersects side $A B$ at point $K$. By Thales' theorem,

$E K = 1 / 3 B E = 1 / 3 A E$. By the same theorem, $A M: M D = A E: E K = 3: 1$.

## Answer

$3: 1$, measured from vertex $A$.