Let $AC=BC=1, BD=CD=\frac{1}{2},AD=\frac{\sqrt 5}{2}$. $AB=\sqrt 2$ Let $F$ lies on $AB$ & $BF=\frac{\sqrt 2}{3}.O$ is the intersection of $AD$ &$CF$ Now,$\text{cos} \angle BCF$ $=\frac{CA}{AB}=\frac{2}{\sqrt 5}$ Apply stewart's theorem as $CF=d$, as a chevian, $AC^2\cdot BF+BC^2\cdot AF=AB(d^2+AF\cdot BF)$ $\implies d=\frac{\sqrt 5}{3}$. In $\triangle BCF$ apply law of cosine and get $\text {cos}\angle BCF=\frac{2}{\sqrt 5}$ $\implies \text {cos}BCF=\frac{2}{\sqrt 5}=\text{cos} \angle CAB$ So, $\triangle DAO$ is similar to $\triangle ACD$. So, $AD\perp CF$. But we know $AD\perp CE$. So, $F=E$. $\implies AE=2BE$

We have: $0 = \overrightarrow{AD} . \overrightarrow{CE} = \dfrac{1}{2} (\overrightarrow{AB} + \overrightarrow{AC})(\overrightarrow{CB} + \overrightarrow{BE}) = \dfrac{1}{2} (\overrightarrow{AB} . \overrightarrow{CB} + \overrightarrow{AB} . \overrightarrow{BE} + \overrightarrow{AC} . \overrightarrow{CB} + \overrightarrow{CB} . \overrightarrow{BE})$ $= \dfrac{1}{2} \left(BC^2 - AB . BE - \dfrac{BC . BE}{\sqrt{2}}\right)$ Then: $BE = \dfrac{BC^2 \sqrt{2}}{AB \sqrt{2} + BC} = \dfrac{BC^2 \sqrt{2}}{3 BC} = \dfrac{BC \sqrt{2}}{3} = \dfrac{AB}{3}$ or $AE = 2 BE$

Let $M$ be the midpoint of $AB$ and $G$ the intersection point of $AD, CM.$ Obviously $G$ is the centroid of $ABC$ and the orthocenter of $CAE.$ Hence, $EG||BD$ and $\frac{{AE}}{{EB}} = \frac{{AG}}{{GD}} = 2$

$F$ - reflection of $C$ about $AB$, $\{\ G\ \}\in CE\cap BF$, then $E$ is centroid of $\triangle BCF$ ($G$ is midpoint of $BF$), done. Best regards, sunken rock