Can anyone tell me what a "harmonic quadruple" is?
I had a problem in Australian mathematical Olympiad paper. The solution uses something like harmonic quadruple to prove that two sides are parallel.
So can this concept be used to prove the following?
$O$ is the point inside triangle $ABC$. The lines joining the three vertices $A$, $B$, $C$ to $O$ cut the opposite sides in $K$,$L$ and $M$ respectively. A line through $M$ parallel to $KL$ cuts the line $BC$ at $V$ and $AK$ at $W$. Prove that $VM=MW$.
The concept has many names, including harmonic quadruple, harmonic conjugates, harmonic throws, and certainly some more. It's a property about four points on a line (which in some situations may be the complex “line” $\mathbb C^1$), or more specifically two pairs of points, which can be described in various ways. A common one is the following incidence configuration, taken from Wikipedia:
The points $\{\{A,B\},\{C,D\}\}$ form a harmonic quadruple. Another way to express this is via the cross-ratio: harmonic quadruples have cross ratio $-1$ (or $2$ or $\frac12$ depending on the order of the points).
Harmonic quadruples are essential building blocks in many approaches to projective geometry. On the one hand, the configuration above is pure incidence, with no metric information, and as a result any transformation which preserves collinearity will preserve harmonic quadruples. On the other hand, harmonic quadruples can be used to express all elementary arithmetic operations, namely addition, subtraction, multiplication, division, squaring. Not square roots. By expressing I mean that if you fix a projective basis (three points $0$, $1$ and $\infty$) on a line, you can express arithmetic on that line using constructions resulting in harmonic conjugates.
So how does this apply to your situation? Well, apart from incidence information you have two additional concepts, namely parallels ($LK\parallel VW$) and midpoints ($M$ is midpoint of $VW$). But these two can be translated into the language of harmonic conjugates. Parallel lines are lines meeting at infinity. And if $(A,B;C,\infty)=-1$, i.e. these four points are a harmonic quaquadruple, then $C$ is the midpoint of $AB$. You can see this if you imagine moving the point $C$ further and further to the right in the above configuration, and as you do, $MN$ will approach parallelism with $AB$, and $D$ will become the midpoint of $AB$.
So to show that $M$ is the midpoint of $VW$ you need to show $(V,W;M,\infty)=-1$, where $\infty$ is the point of infinity on the line $VW$. Unfortunately the witness construction given above can't be found in your construction straight away. We have to pick a slight detour. There are probably many different alternatives one could pick, but I'll go for this one: Label the point where $AB$ intersects $KL$ as $P$. Then $(A,B;M,P)=-1$ as implied by the witness construction, using $C,K,L,O$ as the other points.
Now I did mention earlier on that harmonic quadruples are preserved by projective transformations. Which implies that a central projection from one line to another will preserve it. Use $K$ as that center of projection, and project points from $AB$ to $VW$. This takes $A$ to $W$, $B$ to $V$, $M$ stays on the intersection and $L$ becomes $\infty$ since $KP\parallel VW$. This leads to $(W,V;M,\infty)=-1$ q.e.d.