Parallel lines from a median and angle bisector in a triangle 
Suppose we have $\triangle DEF$ with $|\overline{DE}| < |\overline{DF}|$. Let the midpoint of $\overline{EF}$ be $G$, and let the bisector of $\angle D$ meet side $\overline{EF}$ at $H$. Let the foot of the perpendicular from $E$ to $\overline{DH}$ be $I$, and extend $\overline{EI}$ to meet median $\overline{DG}$ at $J$. Prove that $\overleftrightarrow{HJ}$ and $\overleftrightarrow{DE}$ are parallel.

I got this problem from a professor and I am not sure how to go about it. It LOOKS parallel, but I can't prove this. I am trying coordinates but it is not working out well.

 A: 
take mid point $M$ of $DE$,connect $IM$,cross $EF$ at $G_1$, sine $\triangle DIE$ is right triangle, $\implies \angle IME= 2\angle IDE=\angle FDE \implies IM // DF \implies G_1=G$
make $HJ' //DE$, cross $IM$ at $K \implies \dfrac{HG}{GE}=\dfrac{HK}{ME}=\dfrac{HJ}{DE}$,(it is trivial that $K$ is mid point of $HJ'$)
connect $DJ'$, cross $EF$ at $G_2 \implies \dfrac{HG_2}{G_2E}=\dfrac{HJ}{DE}=\dfrac{HG}{GE}  \\ \implies G_2=G \implies DG_2=DG  \\ \implies J'=J \implies HJ //DE $
it doesn't matter $EF < DE$.
QED.
A: 
Here's a coordinate proof. Let the trangle's edge-lengths be $d$, $e$, $f$ in the usual arrangement, and assign these coordinates to the vertices:
$$D = (0,0) \qquad E = (f,0) \qquad F = e\,(\cos D, \sin D)$$
Midpoint $G$'s coordinates are easy:
$$G = \frac{1}{2}(E+F) = \left( \frac{f + e \cos D}{2}, \frac{e \sin D}{2} \right)$$
By the Angle Bisector Theorem, we know that $H$ is a point such that $e\,|\overline{HE}| = f\,|\overline{HF}|$, which allows us to write
$$H = \frac{1}{f+e}\left(\;e E + f F\;\right) = \frac{ef}{e+f}\left(\;1 + \cos D, \sin D \;\right)$$
Then we have
$$\begin{align}
\overleftrightarrow{EI}&:\quad x ( 1+ \cos D) + \sin D y = f ( 1 + \cos D) \\
\overleftrightarrow{DG}&:\quad e \sin D x = y ( f + e \cos D )
\end{align}$$
so that
$$J = \left(\frac{f (f + e \cos D)}{e + f}, \frac{e f \sin D}{e + f} \right)$$
which has the same $y$-coordinate as $H$ (but a distinct $x$-coordinate), so that $\overleftrightarrow{HJ}$ is parallel to the $x$-axis, and thus $\overleftrightarrow{DE}$. $\square$

Notes 


*

*At no time does this argument assume $f > e$. We only require that $f \neq e$, so that $H$ and $J$ are distinct points (otherwise $\overleftrightarrow{HJ}$ is undefined).

*If we construct $H^\prime$ as the point where the external bisector at $D$ meets $\overleftrightarrow{EF}$, then we can construct companion points $I^\prime$ and $J^\prime$ as before, to get $\overleftrightarrow{H^\prime J^\prime}\parallel\overleftrightarrow{DE}$. The argument is just as above, except we replace $e$ with $-e$ in the definition of $H^\prime$ (and calculation of $J^\prime$). In this case, the requirement $f \neq e$ guarantees that $H^\prime$ is not a point at infinity.
A: Note. This solution has been revised to simplify part of the argument.
This solution uses the fact that the cross-ratio of four points is invariant under central projection from one line to another.
Let $K$ be the reflection of $E$ with respect to line $DI$. Then triangle $DEK$ is isosceles, $I$ is the midpoint of $EK$, and $K$ is on $DF$ (since the angle at $D$ is bisected by $DI$). We have $K \ne F$ since $DK = DE \ne DF$, hence $H \ne J$. 
Henceforth all ratios of lengths should be considered signed.
My first claim is that $\frac{IH}{ID} = \frac{GH}{GF}$. To see this, note that $G$ and $I$ are the midpoints on the sides $EF$ and $EK$ of triangle $EFK$. It follows that $GI$ is parallel to line $KF$, which is also $FD$. This proves the first claim.
My next claim is that $\frac{IJ}{IE} = \frac{GH}{GF}$. This will be enough since $\frac{IJ}{IE} = \frac{IH}{ID}$ implies that $HJ$ and $DE$ are parallel. To prove the claim, we project line $EF$ onto $EK$ centrally from $D$. The images of $E, G, H, F$ under the projection are $E, J, I, K$, respectively. Since cross-ratios are preserved, we have (bearing in mind that $G$ is the midpoint of $EF$ and $I$ the midpoint of $EK$):
$$\begin{align*}
\frac{KI}{KE} \div \frac{JI}{JE} &= \frac{FH}{FE} \div \frac{GH}{GE}, \\
\frac{1}{2}\frac{JE}{JI} &= \frac{1}{2} \frac{FH}{GH}.
\end{align*}
$$
The last claim follows easily from this and completes the proof.
