Let $H_a$ be the foot of the height passing by $A$. We would like to show that $DC,BE,AH_a$ concur. In order to do this, we will use the trigonometric version of Ceva's theorem. We must prove:
$$\frac{\sin{\angle{EBA}}}{\sin{\angle{EBC}}} \cdot \frac{\sin{\angle{DCB}}}{\sin{\angle{DCA}}} \cdot \frac{\sin{\angle{H_aAC}}}{\sin{\angle{H_aAB}}} = 1$$
After some simple angle chasing, this is equivalent to
$$\frac{\sin{20}}{\sin{40}} \cdot \frac{\sin{70}}{\sin{10}} \cdot \frac{\sin{10}}{\sin{30}} = 1$$
$\iff \sin{20} \sin{70} = \sin{30} \sin{40}$
Using the fact that $\sin 70 = \cos 20$, this is equivalent to
$\frac{1}{2} \sin{40} = \sin{30} \sin{40} \iff \sin{30} = \frac{1}{2}$, which is true.