My solution : Let $ \tau $ be the tangent of $ \odot (ACD) $ through $ A $ . Let $ M $ be the midpoint of $ CD $ and $ T \equiv BC \cap DE $ . From $ \angle ACB=\angle ADM, \angle CBA=\angle DMA \Longrightarrow \triangle ABC \sim \triangle AMD $ . Similarly, $ \triangle AED \sim \triangle AMC \Longrightarrow \angle BAE=2\angle CAD =180^{\circ}-\angle ETB \Longrightarrow T \in \odot (ABE) $ . Since $ AT $ is the isogonal conjugate of $ AM $ WRT $ \angle CAD $ (well-known) , so $ \measuredangle (\tau, AT)=\measuredangle (\tau, AC) +\measuredangle CAT=\measuredangle ADC+\measuredangle MAD=\measuredangle AMC=\measuredangle AET $ , hence $ \tau $ is also the tangent of $ \odot (ABE) $ through $ A \Longrightarrow \odot (ABE) $ and $ \odot (ACD) $ are tangent to each other . Q.E.D

My solution: Let $M$ is the midpoint of $CD\Rightarrow \odot (ABC)\cap \odot (ADE)=\left \{ A,M \right \}$. We have $\angle DME+\angle DEM=\angle xDC=\angle CAD=\angle CAM+\angle MAD=\angle CBM+\angle DEM\Rightarrow \angle DME=\angle CBM$. Similarly $\angle DEM=\angle CMB$. So $\triangle MBC\sim \triangle EMB$ (a.a) $\Rightarrow \frac{MB}{ME}=\frac{MC}{DE}=\frac{MD}{DE}$, and $\angle BME=180^{\circ}-(\angle BMC+\angle DME)=180^{\circ}-(\angle MED+\angle DME)=\angle MDE$ $\Rightarrow \triangle BME\sim \triangle MDE$(s.a.s) $\Rightarrow \angle BEM=\angle MED=\angle BMC$ $(*)$ Let $At$ be the tangent line of $\odot (ACD)$, we have $\angle tAC=\angle ADC\Leftrightarrow \angle tAB+\angle BAC=\angle AEM=\angle AEB+\angle BEM=\angle AEB+\angle BMC$ ($\because (*)$)$=\angle AEB+\angle BAC$ $\Leftrightarrow \angle tAB=\angle AEB$ $\Rightarrow At$ is tangent to $\odot (ABE)$ $\Rightarrow \odot (ACD)$ is tangent to $\odot (ABE)$. DONE Remark: Let $BC\cap DE=F$ then we prove that $M$ is the incenter of $\triangle BEF$, and combine with $\odot (ACD)$ is tangent to $FB, FE$ $\Rightarrow \odot (ACD)$ is the F-mixtilinear incircles of $\triangle BEF$ $\Rightarrow \odot (ACD)$ is tangent to $\odot (ABE)$.

Attachments:

Angle chasing... Let $BC\cap DE=R$ and midpoint of $CD$ is $M$. Then $AR$ is symmedian of $\angle ACD$. Then we have $\angle CAB=\angle CMB=\angle RCD-\angle CBM=\angle CAD-\angle CAM=\angle MAD$, similarly $\angle DAE=\angle CAM$, which means that $\angle BAE=2\angle CAD$ adding to this $\angle BRE= 180-2\angle CAD$ gives that $(R B A E)$ is cyclic. Since $\angle DAM=\angle RAC$ (because $AR$ is symmedian) we have $\angle MEB=\angle DEB-\angle DEM=\angle RAB-\angle DAM= \angle DAM$. Then let $AA$ be the tangent to $(A C D)$ and $K$ be the point on $AA$, then $KAB=\angle KAC-\angle BAC=\angle MDA-\angle MAD=\angle MEA-\angle MEB=\angle BEA$, which completes the solution.