How prove that $AB>AC$ in triangle $ABC$?

Question

Point $D$ is chosen inside $\triangle ABC$, and point $E$ on segment $BD$ such that $BD=CE$. Suppose $\angle ABD=\angle ECD=10^{\circ}$, $\angle BAD=40^{\circ}$, and $\angle CED=60^{\circ}$.How prove that $AB>AC$? I have no idea how to do this, can this be proved with simple geometry?

Answer 1

Hint: Draw the isosceles triangle $\triangle AFD$ where $F$ is on $AB$, which has common angle $70^\circ$. Find two congruent triangles. Use triangle inequality.