# Geometry problem from a Berkeley course

I've been trying to solve this problem proposed as part of one of the first lectures of a Berkeley linear algebra course:

"What Good is a Basis ? The freedom to choose a basis often simplifies calculations and proofs. For instance, here is a phenomenon first noticed by G. Desargues (1593 - 1662), a contemporary of R. Descartes: In the plane, fix two intersecting straight lines B and C and a point p on neither. Through p draw two straight lines X and Y that intersect B and C in four points all told. Two pairs of those points are not yet joined by straight lines; draw those lines now and, if they intersect, call their intersection q . As X and Y move, always passing through p , so does q move; show that it moves along some fixed straight line D . To prove the existence of D takes some ingenuity if none but the methods of Euclidean plane geometry may be used; and if rectangular Cartesian coordinates must be used the proof is a tedious computation. But a relatively short computation suffices if we choose an apt basis. Test these claims by trying to verify Desargues’ observation above using only the ideas you learned in High-School. Then you will be better able to appreciate the strategy motivating vector notation and its algebra used in the following proof."

The construction suggested reminds me of concepts like Ceva's theorem, complete quadrangle, internal/external division, which I learned about in Geometria Moderna (Modern Geometry required subject for math majors at UNAM Mexico). Sadly, though, I can't think of a way to use any of those concepts to solve the problem. Suggestions, please!

• Ah, Velvel Kahan. His stuff's always nice... – J. M. isn't a mathematician Dec 19 '11 at 14:05
• Care to include the actual question? – Did Dec 19 '11 at 14:08
• Relying completely on an off-site link is discouraged, but referring to a "problem proposed on a page" when the page doesn't say "Problem:" anywhere and we'd have to read through it to understand what you're trying to ask is taking non-selfcontainedness one step further. Please take into account that any effort you spend in presenting the question efficiently is spent only once and any effort you leave to the reader is spent dozens or hundreds of times. – joriki Dec 19 '11 at 14:26
• @joriki In addition, some people (me included) might ignore to even look up the "problem" at all. – AD. Dec 19 '11 at 15:33
• I did (ignore to look up). – Did Dec 19 '11 at 17:03

I choose a different labelling:
Let $A$,$B$,$C$,$D$ be 4 distinct points in the plane
Let $X = AB \cap CD$
Let $Y = AC \cap BD$
Let $Z = AD \cap BC$
Let $P = AD \cap XY$
Let $Q = BC \cap XY$
Let $R = AC \cap XZ$
Then we have cross-ratios $(A,C;Y,R) = (A,D;P,Z) = (C,B;Q,Z) = (C,A;Y,R) = 1/(A,C;Y,R)$
Thus $(A,C;Y,R) = -1$ (Alternatively, use Ceva's theorem and Menelaus' Theorem)
Thus if we fix $A$,$C$,$X$,$Y$ and let $BD$ vary, $R$ is fixed and $Z$ must be on $XR$
Thus $XZ$ remains the same line
By symmetry we can subsequently fix $B$,$D$,$X$,$Y$ and let $AC$ vary, $XZ$ remains the same line
Thus $Z$ always remains on the same line
(QED)

Note that the invariance of cross-ratios under projection from line to line can be proven using Menelaus' theorem twice. Cross-ratios are also invariant under projection from circle to line through a point on the circle and this can be proven using inversion.

• Thank you so much, very nicely written and very informative. – Miguel Morales Jan 11 '12 at 5:16