Problem
GEO-B2-M09-P023 Newton Line
In quadrilateral \(ABCD\), lines \(AB\) and \(CD\) meet at \(E\), while \(AD\) and \(BC\) meet at \(F\); assume these intersections are finite. Prove by coordinates that the midpoints of \(AC\), \(BD\), and \(EF\) lie on one line.
Take \(E\) as the origin, and use lines \(AB\) and \(CD\) as coordinate axes.
Choose affine coordinates so that \(E=(0,0)\), \(A=(a,0)\), \(B=(b,0)\), \(D=(0,d)\), \(C=(0,c)\). Let \(F=(r,s)\). Since \(F\in AD\), we have \(\frac{r}{a}+\frac{s}{d}=1\), so \(rd+as=ad\). Since \(F\in BC\), we have \(\frac{r}{b}+\frac{s}{c}=1\), so \(rc+bs=bc\). The midpoints of \(AC\), \(BD\), \(EF\) have coordinates \(\left(\frac a2,\frac c2\right)\), \(\left(\frac b2,\frac d2\right)\), \(\left(\frac r2,\frac s2\right)\). It remains to prove that \((r,s)\) lies on the line through \((a,c)\) and \((b,d)\). The equation of this line gives the expression \((d-c)r-(b-a)s-a(d-c)+c(b-a)\). It equals \((rd+as-ad)-(rc+bs-bc)=0\). Hence the three midpoints are collinear.
A strong problem on affine coordinates; it is useful to discuss why axes may be chosen along two intersecting lines.