$AE \ge h$ and $BC=BE+CE \ge 2\sqrt{BE \cdot CE}$ $\Longrightarrow$ $BC^2 \ge 4 \cdot BE \cdot CE$ $\Longrightarrow \frac{4h}{BC} \cdot CE \le \frac{AE}{BE} \cdot BC=\frac{CD}{BC} \cdot BC=CD.$ Equality obviously holds when $AE=h,$ $BE=CE,$ i.e. $E$ is midpoint of $\overline{BC}.$

In a triangle $ABC$ assume $AB = AC$, and let $D$ and $E$ be points on the extension of segment $BA$ beyond $A$ and on the segment $BC$, respectively, such that the lines $CD$ and $AE$ are parallel. Prove $CD \ge \frac{4h}{BC}CE$, where $h$ is the height from $A$ in triangle $ABC$. When does equality hold?