Prove three chords of a circle are concurrent iff their poles with respect to a circle are collinear. This probably would be a very simple problem if I could use any theorem I wanted about poles and polars, but in the book they give a definition and they say the problem should be solved using only that:
"Let C be a circumference and w a chord, the pole P of w is the intersection of the tangents through the points of intersection of w with C."
Thanks for any help.
 A: Let us draw the picture first, fix notations and make a more precise claim:

Let $(O)$ be a circle with center $O$. Let $A_1A_2$, $B_1B_2$, $C_1C_2$ be chords of the circle $(O)$ intersecting in $X$. Let $M,N,P$ be respectively the mid points of these chords. Let $S\to S^*$ be the inversion centered in $O$ that fixes the circle $(O)$. Let $M^*$, $N^*$, $P^*$, $X^*$ be the images of $M,N,P,X$ by this inversion.

Then the points $M^*$, $N^*$, $P^*$, $X^*$ are colinear, and $OXX^*$ is perpendicular in $X^*$ on the corresponding line of colinearity.
Reciprocally, if we do not assume the concurence of the three chords, (so there is no point $X$ in the picture,) but assume the linearity of $M^*,N^*, P^*$, then $M,N,P$ are on a circle through $P$, with diameter supported by the perpendicular from $O$ on $M^*N^*P^*$.

Proof:
The inversion maps circles through $O$ in lines. If $OX$ is the diameter of such a circle, the transformed map of the circle is a line perpendicular in $X^*$ on $OX^*$.
Comment: In our case, the perpediculars in $A_1,A_2$ on $OA_1$, respectively $OA_2$, meet in a point $A$, say for a short time, so that $OA_1^2=OA_2^2=OM\cdot OA$. So $A$ is exactly the pole $M^*$.
$\square$
