Problem: Given a right triangle $ABC$ with $A=90°$. On $AC$ we label $D$ such that $\angle ABD=(1/3)\angle ABC$. On AB we label $E$ such that $\angle ACE=(1/3)\angle ACB$. $F$ is the intersection of $BD$ and $CE$. The angle bisector of BFC and FBC meet at $I$. Prove that $DIE$ is an isosceles triangle.
So far I've find out that $\angle BFC=120°$, so $\angle BFI=\angle IFC=60°$. I cannot figure out any pairs of equal triangle.
Note: I have learnt equal triangle. But I haven't learnt trigonometry, of similar triangle or any property of circle or quadrilateral.