Let be $I$ the incenter of the triangle with the measure of $\angle A = 90$ degrees. If point $D$ lies on line $AB$ and $E$ lies on line $AC$ such that $BD \cong CE \cong BC $ and lines $BE$ and $CD$ intersect each other in $F$, then prove that $FEID$ is a parallelogram.
I know there's a few basic ways to prove a quadrilateral is a parallelogram, such as showing opposite sides are congruent or parallel. But I'm not too sure how to make use of all these conditions though. There seems to be a lot of different information and I'm not quite sure how to make use of all of it. Any ideas?
Just prove by angle chasing that two pairs of opposite angles are equal in $FEID$ and thereafter it's a parallelogram.
It turns out that $\angle EFD=\angle EID=45^{\circ}$ and $\angle FEI=\angle FDI=135^{\circ}$.