In triangle $ABC$, $\angle ACB=50^{\circ}$ and $\angle CBA=70^{\circ}$. Let $D$ be a foot of perpendicular from point $A$ to side $BC$, $O$ circumcenter of $ABC$ and $E$ antipode of $A$ in circumcircle $ABC$. Find $\angle DAE$
1. **Determine the missing angle in triangle $ABC$:**
   Given $\angle ACB = 50^\circ$ and $\angle CBA = 70^\circ$, we can find $\angle BAC$ using the fact that the sum of angles in a triangle is $180^\circ$.
   \[
   \angle BAC = 180^\circ - \angle ACB - \angle CBA = 180^\circ - 50^\circ - 70^\circ = 60^\circ
   \]

2. **Identify the properties of the circumcircle and the antipode:**
   - The circumcenter $O$ is the center of the circumcircle of $\triangle ABC$.
   - The antipode $E$ of $A$ in the circumcircle is the point such that $AE$ is a diameter of the circumcircle. Therefore, $\angle AOE = 180^\circ$.

3. **Use the isogonal conjugate property:**
   - $AD$ is the altitude from $A$ to $BC$, making $\angle ADB = 90^\circ$.
   - $AE$ is the diameter of the circumcircle, making $\angle AEC = 90^\circ$.
   - Since $AD$ and $AE$ are isogonal lines with respect to $\angle BAC$, we have $\angle BAD = \angle EAC$.

4. **Calculate $\angle EAC$:**
   - Since $E$ is the antipode of $A$, $\angle EAC$ is half of $\angle BAC$ because $E$ lies on the circumcircle opposite to $A$.
   \[
   \angle EAC = \frac{1}{2} \angle BAC = \frac{1}{2} \times 60^\circ = 30^\circ
   \]

5. **Determine $\angle DAE$:**
   - Since $\angle BAD = \angle EAC$, we have:
   \[
   \angle BAD = 30^\circ
   \]
   - Therefore, $\angle DAE = \angle BAD = 30^\circ$.

The final answer is $\boxed{30^\circ}$