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In a quadrilateral $ABCD$, let $AB$ and $CD$ meet at $E$ and $AD$ and $BC$ meet at $F$ then prove that the midpoints of $AC, BD$ and $EF$ are collinear

Following is a geogebra diagram. Here $H$ is the midpoint of $BD$, $I$ is the midpoint of $AC$ and $G$ is the midpoint of $EF$ problem diagram

What I tried to do was label all the points where $IH$ meets other points and tried to prove that $\frac{EJ}{JB}.\frac{BK}{KF} = -1$ so that by multiplying $\frac{FG}{GE} = 1$ you could prove that $K,J,G$ are collinear by the Menelaus theorem and consequently $I,H,G$ are collinear. However this brute-force method did not work. So how do I solve this?

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You may assume $A=(0,1)$, $B=(0,0)$, $C=(1,0)$, $D=(u,v)$ with $u>0$, $v>0$, $u+v>1$. The remaining points are now easy to compute.

My suspicion is that the claim is a well known theorem in projective geometry, having to do with "harmonic point pairs".

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  • $\begingroup$ Yes, its related to Desargues' theorem, till what I have found. $\endgroup$ – Helix Apr 5 '18 at 7:54
  • $\begingroup$ Would'nt assuming that $A = (0,1), B = (0,0), C = (1,0)$ also be assuming that $\angle ABC = 90^\circ$? $\endgroup$ – Helix Apr 5 '18 at 8:00
  • $\begingroup$ @Helix: No. It just means that I have chosen $\vec{BA}$ and $\vec{BC}$ as a basis in the sense of linear algebra. There are no angles in the argument, just lines, intersections, and midpoints. $\endgroup$ – Christian Blatter Apr 5 '18 at 8:05
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Let $\vec{CE}=\vec{a}$, $\vec{CF}=\vec{b}$, $\vec{CD}=x\vec{a}$ and $\vec{CB}=y\vec{b}.$

Thus, $$\vec{CG}=\frac{1}{2}(\vec{a}+\vec{b})$$ and $$\vec{CH}=\frac{1}{2}(x\vec{a}+y\vec{b}).$$ Now, $$\vec{FA}=k\vec{FD}=k(x\vec{a}-\vec{b})$$ and $$\vec{AB}=m\vec{EB}=m(y\vec{b}-\vec{a}).$$ Thus, since $$\vec{FB}=\vec{FA}+\vec{AB},$$ we obtain: $$-(1-y)\vec{b}=k(x\vec{a}-\vec{b})+m(y\vec{b}-\vec{a}),$$ which gives $k=\frac{1-y}{1-xy}$ and $m=\frac{x(1-y)}{1-xy}.$

Hence, $$\vec{FA}=\frac{1-y}{1-xy}(-\vec{b}+x\vec{a})$$ and $$\vec{CI}=\frac{1}{2}\left(\vec{b}+\frac{1-y}{1-xy}(-\vec{b}+x\vec{a})\right)=\frac{1}{2}\left(\frac{x(1-y)}{1-xy}\vec{a}+\frac{y(1-x)}{1-xy}\vec{b}\right).$$ Now, $$\vec{IH}=\vec{CH}-\vec{CI}=\frac{xy}{2(1-xy)}((1-x)\vec{a}+(1-y)\vec{b}),$$ $$\vec{IG}=\vec{CG}-\vec{CI}=\frac{1}{2}((1-x)\vec{a}+(1-y)\vec{b})$$ and we are done!

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I think you can find a solution here:

http://forumgeom.fau.edu/FG2012volume12/FG201212.pdf

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  • $\begingroup$ This document just gives some properties of the line, it dose not actually prove its existence. $\endgroup$ – Helix Apr 5 '18 at 15:06
  • $\begingroup$ Sorry, didn't actualy read it $\endgroup$ – Aqua Apr 5 '18 at 15:24
  • $\begingroup$ Its okay, I searched up the Newton-Gauss line and found a proof for it, so knowing its name helped. $\endgroup$ – Helix Apr 5 '18 at 15:29

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