Problem
GEO-B3-M02-P020 Harmony in a Complete Quadrangle
B. New Original Problem. Let \(A,B,C,D\) be four points, no three collinear. Define \(E=AB\cap CD\), \(F=AD\cap BC\), and \(G=AC\cap BD\). The line \(EF\) meets \(AC\) at \(H\). Prove that \((A C G H)=-1\).
C. Hint 1. This is the classical harmonic fact of a complete quadrangle.
D. Hint 2. Projectively reduce the quadrangle to a convenient coordinate model.
E. Full Solution.
Cross-ratio is preserved by projective transformations, so we may replace the complete quadrangle by a projectively equivalent convenient model. Choose coordinates so that
\[ A=(-1,0),\quad C=(1,0),\quad B=(0,1),\quad D=(0,-1). \]
Then \(G=AC\cap BD=(0,0)\). The lines \(AB\) and \(CD\) have the same direction, so their intersection \(E\) is the point at infinity of direction with slope \(1\). The lines \(AD\) and \(BC\) are also parallel, so \(F\) is the point at infinity of direction with slope \(-1\).
Therefore the line \(EF\) is the line at infinity, and its intersection with \(AC\) is the point at infinity of the line \(AC\). Thus, in the coordinate on \(AC\), we have \(A=-1\), \(C=1\), \(G=0\), and \(H=\infty\).
Hence
\[ (A C G H)=(-1,1,0,\infty)=\frac{0-(-1)}{0-1}=-1. \]
Since this cross-ratio is preserved by the inverse projective transformation, the original configuration also satisfies \((A C G H)=-1\).
The coordinate model makes the harmonic relation transparent: two pairs of opposite sides become parallel, and \(H\) goes to infinity.