# Prove midpoints collinear

Let $ABCD$ be a convex quadrilateral and let $E$ and $F$ be the points of intersections of the lines $AB, CD$ and $AD,BC$ , respectively. Prove that the midpoints of the segments $AC$, $BD$, and $EF$ are collinear.

I tried to solve this question

assuming the opposite edges aren't parallel Let G,H, and I are midpoint of BD, AC, EF, respectively. Thus we have $[AGB]+[CGD]=\frac{1}{2}([ABD]+[BCD])=\frac{1}{2}[ABCD]$ similarity, $[AHB]+[CHD]=\frac{1}{2}([ABC]+[ACD])=\frac{1}{2}[ABCD]$ I can only do until here. I can't prove $G,H,I$ collinear. Could you help me continue my works? Thank you :D

• You're assuming the opposite edges aren't parallel. What have you tried? – Ted Shifrin Jul 24 '13 at 12:45
• Another problem from this poster with no source, no motivation, and no indication of the slightest bit of effort. That's not what this website is here for. – Gerry Myerson Jul 24 '13 at 12:54
• i'm sorry, Mr. Gerry Myerson. I'm a new user here and I don't know anything about the rules here. If i make a mistake, i'm so sorry. Okay, i'll add my works in this question.. – Rizky Darmawan Brekele Jul 24 '13 at 13:05

## 1 Answer

Hint: Call $\overrightarrow{AB} = \vec x$ and $\overrightarrow{AD} = \vec y$. Find expressions for $\overrightarrow{AG}$, $\overrightarrow{AH}$, and $\overrightarrow{AI}$ as linear combinations of $\vec x$ and $\vec y$. You will need to use all the geometry in the problem, for example, writing $\overrightarrow{AE} = s\overrightarrow{AB}$ for some scalar $s$, etc.