Think circularly

$AD= BD = R, AE = BE$ $\angle BAE = 10^\circ , \angle BAD = 20^\circ , \angle ADB = 140^\circ$

By Inscribed Angle Theorem $\angle ACB= 140^\circ / 2 = 70^\circ$ .

As $D$ is circumcenter of $\triangle ABC$ $\Rightarrow$ $BD=AD$ and $AE=BE$ .

As $\triangle AEB$ is isosceles, $\angle BAE=\angle ABE= \frac{180-160}{2}=10°$

$BE$ and $AE$ are bisectors of $\angle ABD$ and $\angle BAD$ respectively $\Rightarrow$ $\angle ABD=\angle BAD=20°$

Then $\angle BDA=180-40=140°$

Finally, as $D$ is circumcenter by inscribed angle $\angle ACB=\frac{\angle BDA}{2}=\boxed{70°}$

As AE & BE are the two angle bisectors of Triangle ABD, we can state that,

In Triangle ABE, (1/2)A+(1/2)B+E = 180 ............. (a).

In Triangle ABD, A+B+D = 180; That implies , (1/2)A+(1/2)B+(1/2)D = 90 ........... (b)

Solving the (a) & (b) no. equation, We get, The angle AEB = (90 Degree + (1/2) Angle ADB). As The Angle AEB = 160 Degree, So, The {(1/2) Angle ADB} equals to 70 Degree. Because D is the circumcenter of Triangle ABC, so, Angle ACB must equals to {(1/2) Angle ADB}. Therefore, the answer is Angle ACB = 70 Degrees.

Just almost the same solution.