# 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? 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. 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.. Jul 24 '13 at 13:05
• This can also be solved by simple POP Aug 10 '20 at 2:36

## 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.