MathNet 2018-part-04 — 181 problems (EN–VI) Question 179

Let $\triangle ABC$ be a right triangle in $A$ and points $D$ and $E$ on $AB$ such that $\angle ACD = \angle DCE = \angle ECB$. Show that if $3\overrightarrow{AD} = 2\overrightarrow{DE}$ and $\overrightarrow{CD} + \overrightarrow{CE} = 2\overrightarrow{CM}$, then $\overrightarrow{AB} = 4\overrightarrow{AM}$.
Gabriel Popa