Hahahaha..... It's also very simple problem.Answer is $ 180-arccos\frac{1}{2\sqrt{2}} $.

Sardor. You are wrong. PLease! Make attention your solution again!

Let $BD=3k,CE=2k$.Then by the power of point theorem we have $3k(3k+5)=4(4+2k) \Rightarrow (9k+16)(k-1)=0 \Rightarrow k=1 \Rightarrow \boxed{BD=3,CE=2}$ $\triangle{BDE} \sim \triangle{BCA} \Rightarrow \frac{DE}{AC}=\frac{BD}{BC}=\frac{1}{2} \Rightarrow \boxed{DE=\sqrt{7}}$.Now, $\frac{AD^2+(2\sqrt{7})^2-CD^2}{2*5*2\sqrt{7}}=\cos{\angle{DAC}}=\cos{\angle{DEB}}=\frac{\sqrt{7}}{4} \Rightarrow \boxed{DC=3\sqrt{2}}$ Finally $\cos{\angle{BDC}}=\frac{BD^2+CD^2-BC^2}{2BD*CD}=\frac{9+18-36}{18\sqrt{2}}=-\frac{1}{2\sqrt{2}} \Rightarrow \angle{BDC}=180^{\circ}-\arccos(-\frac{1}{2\sqrt{2}})$. Uzbekistan,can you find out the fault in my proof.

I think everything you have up to $\cos{(\angle BDC)} = -\frac{1}{2\sqrt{2}}$ is right. Then $\angle BDC$ should just equal $\arccos{\left(-\frac{1}{2\sqrt{2}}\right)}$ which is equal to $180 - \arccos{\left(\frac{1}{2\sqrt{2}}\right)}$, agreeing with Sardor's edited answer and my answer.

Sardor wrote: Hahahaha..... It's also very simple problem.Answer is $ 180-arccos\frac{1}{2\sqrt{2}} $. what is mean? Rad (radian) or $ ^\circ $ ?

Obviously $ ^o $