Let I,I_a be incenter and excenter of ABC. ADEI_a and BIOCI_a are cyclic. EDI_a=I_aAD=I_aAC+ACI=I_aIC=I_aBC=>(I_aDE), (IBC) are homothetic.

Let $I_A,I_B,I_C$ be the excenters of triangle $ABC$. Let $I$ be the incenter. We will show that $\odot (BIC) , \odot (ADE)$ are tangent. Obviously, $A,D,E,I_A$ are cyclic and so are $B,I,C,I_A$. Now the centers of these circles are obviously the mid points of $AI_A, II_A$. As $A,I,I_A$ are collinear, the line joining the centers of the two circles $\odot (ADE), \odot (BIC)$ Lies on the common point $I_A$, which means they are tangent at $I_A$. As $B,I,O,C$ are cyclic, we conclude that $\odot (BOC), \odot (ADE)$ are tangent. $\blacksquare$.

Let $T=BD\cap CE$. Note that $\angle DBA=90^{\circ}-\frac{\angle B}{2}$ and $\angle ACE= 90^{\circ}-\frac{\angle C}{2}$ so $\angle DAE=\angle DAB+\angle CAE +\angle BAC=(90^{\circ}-\angle DBA)+(90^{\circ}-\angle ACE)+60^{\circ}=(\frac{\angle B}{2}+\frac{\angle C}{2})+60^{\circ}=120^{\circ}.$ Also, note that $\angle BOC=2\angle A=120^{\circ}.$ Because $\angle ADT+\angle AET=180^{\circ}$, $ADTE$ is cyclic. Hence, $\angle DTE=180-\angle DAE=60^{\circ}$. By construction, $\angle BTC=\angle DTE=60^{\circ}$ so $OBTC$ is cyclic. Therefore, $T$ is the common tangency point of $(ADE)$ and $(BOC)$, as desired. $\blacksquare$

Let $DB\cap EC=X$ and circumcenter of $OBC$ be $O'$. $\bullet$ $ADXE$ is cyclic. $\bullet$ $OBXC$ is cyclic. $\bullet$ $ABO'C$ is cyclic. $\bullet$ $\triangle OBO'$ and $\triangle OCO'$ are equialteral. Using these and doing some angle chasing we can easily find that $A-O'-X$ are collinear. It means that circumcircles of $\triangle ADE$ and $\triangle OBC$ touches at $X$ because, centers of these circles and $X$ are collinear.

Let DB and EC meet at S. ∠ADS = ∠AES = 90 ---> ADSE is cyclic. ∠BSC = 60 and ∠BOC = 2∠BAC = 120 ---> BOCS is cyclic. OB and AD are both perpendicular to DB and OC and AE are both perpendicular to CE and ∠BOC = ∠DAE = 120 ---> triangles DAE and BOC are homothetic so S is their tangency point.