# 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.

• As @achillehui's diagram shows, the result is true even if $|\overline{DE}| > |\overline{DF}|$. What's really important is that $|\overline{DE}|\neq|\overline{DF}|$, so that $G$ and $H$ don't coincide. (BTW: I have a coordinate proof of the result, but am seeking a cleaner approach.) – Blue Jan 4 '16 at 3:25
• Could I see the coordinate proof? I actually was attempting coordinates but it was not working out. – user298322 Jan 4 '16 at 3:32
• @Blue, oops, I draw the wrong figure. I didn't notice that b/c the statement also looks true in that case. – achille hui Jan 4 '16 at 3:38
• This question is very similar to Problem 4 of the current round of the USAMTS (an ongoing contest). The deadline for submitting solutions is 3PM EST on January 4th. – JimmyK4542 Jan 4 '16 at 5:12
• Deleting my answer until after the USAMTS deadline. (Note that the original problem statement is identical to the contest problem, except for the naming of points. I edited it into the current form without realizing the source.) – Blue Jan 4 '16 at 6:14

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.

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.

• What is the ratio IJ to JE in terms of points D,E,F – Kshitij Singh Jan 14 '19 at 17:33
• @KshitijSingh: I don't have time to revisit a three(!)-year-old problem right now, but the ratio you seek should be fairly straightforward to determine from the information I've given. All you need is to determine the coordinates of point $I$. You might take a look at the other two answers, which avoid coordinates and feature ratios. – Blue Jan 14 '19 at 18:15

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.

• The main thing to see here is that once $K$ (the reflection of $E$ with respect to $I$) is placed in the figure, all the other points can be determined in an affinely invariant way in terms of the triangle $DEK$ and the choice of a point $F$ on side $DK$. What I mean by this is that $G, H, I, J$ can be defined solely in terms of midpoints of segments and intersections of lines, and these constructions are invariant under affine transformations. From that point forward, $DEK$ can be replaced with any other convenient triangle. It becomes plausible that a purely projective proof is possible.... – David Jan 4 '16 at 21:04
• ...Alternatively, let $D= (0,0)$, $E=(1,0)$, $K = (0,1)$, $F = (0,c)$ and use coordinates. For the reasons stated, it makes no difference that the coordinate system is not rectangular, so long as $I$ is defined as the midpoint of $EK$. – David Jan 4 '16 at 21:06