Prove that secants of a circle pass through a common point Let $A$, $B$, $C$, $D$, and $E$ be five points on a circle. For any three points, we draw the line going through the centroid of the triangle formed by these three points that is perpendicular to the line passing through the other two points. (For example, we draw the line going through the centroid of triangle $BDE$ that is perpendicular to $AC$.) In this way, we draw a total of $ \binom{5}{3}= 10 $ lines. Show that all $10$ lines pass through the same point.
 A: One can find the answer in a straightforward way by determined two sample lines, computing their intersection, and noting that the result is symmetric in the five original points.
One can argue somewhat more abstractly that, if such a point ---call it $P$--- exists, then its coordinates "should" be a linear combination of those of $A$, $B$, $C$, $D$, $E$, which we take to be on a circle about the origin. By symmetry, the combination has to amount to a simple scalar multiple of the sum ---call it $S$--- of those coordinates:
$$P = p\;(A+B+C+D+E) = p\;S$$
Now, let $X$ and $Y$ be two of the points on the circle, and let $K$ be the centroid of the remaining three points. Then we have
$$K = \frac13(A+B+C+D+E+F-X-Y) = \frac13S-\frac13(X+Y)$$
Since $\overline{PK}\perp\overline{XY}$, the dot products of the corresponding vectors vanish:
$$\begin{align}
0 = (P-K)\cdot(X-Y) &= \left(\left(p-\frac13\right)S+(X+Y)\right)\cdot(X-Y) \\
&= \left(p-\frac13\right)S\cdot(X-Y) + (X+Y)\cdot(X-Y) \tag{$\star$}
\end{align}$$
Now, $(X+Y)\cdot(X-Y) = X\cdot X-Y\cdot Y = |X|^2 - |Y|^2 = 0$, since $X$ and $Y$ both lie on the same origin-centered circle. Consequently, $(\star)$ becomes
$$0 = \left(p-\frac13\right)\;S\cdot(X-Y) \tag{$\star\star$}$$
which must hold for all choices of $X$ and $Y$, despite $S$ being "constant" for all those choices. This can only happen if:


*

*$A$, $B$, $C$, $D$, $E$ all coincide;

*$S = 0$; or

*The scalar multiplier itself is zero: $p = 1/3$.


Discarding $(1)$ as problematic ("the line passing through the other two points" would be ill-defined), we see that we can accommodate both $(2)$ and $(3)$ by taking
$$P = \frac13 S$$
So, not only "should" $P$ be a linear combination of the original points, it is a linear combination of the original points!
