In the triangle $A B C$, let $l$ be the bisector of the external angle at $C$. The line through the midpoint $O$ of the segment $A B$ parallel to $l$ meets the line $A C$ at $E$. Determine $|C E|$, if $|A C|=7$ and $|C B|=4$.
Let $F$ be the intersection point of $l$ and the line $A B$. Since $|A C|>|B C|$, the point $E$ lies on the segment $A C$, and $F$ lies on the ray $A B$. Let the line through $B$ parallel to $A C$ meet $C F$ at $G$. Then the triangles $A F C$ and $B F G$ are similar. Moreover, we have $\angle B G C=\angle B C G$, and hence the triangle $C B G$ is isosceles with $|B C|=|B G|$. Hence $\frac{|F A|}{|F B|}=\frac{|A C|}{|B G|}=\frac{|A C|}{|B C|}=\frac{7}{4}$. Therefore $\frac{|A O|}{|A F|}=\frac{3}{2} / 7=\frac{3}{14}$. Since the triangles $A C F$ and $A E O$ are similar, $\frac{|A E|}{|A C|}=\frac{|A O|}{|A F|}=\frac{3}{14}$, whence $|A E|=\frac{3}{2}$ and $|E C|=\frac{11}{2}$.