Some obviousness about harmonic divisions

by math_explorer, Mar 14, 2011, 10:29 AM

If $A, B, C, D$ are on a line (in that order), then these statements are pairwise equivalent:

- $ABCD$ is a harmonic division
- $B$ and $D$ are inverses w.r.t. the circle with diameter $AC$
- $A$ and $C$ are inverses w.r.t. the circle with diameter $BD$
- the circle with diameter $AC$ is a circle of Apollonius with foci $B, D$
- the circle with diameter $BD$ is a circle of Apollonius with foci $A, C$

If $O$ is a point outside $ABCD$:

$\angle AOC$ is right iff $\angle BOD$ is bisected (internally/externally) by $OA$ and/or $OC$;
$\angle BOD$ is right iff $\angle AOC$ is bisected (internally/externally) by $OB$ and/or $OD$;

$\frac{AB}{BC} = \frac{AD}{DC}$
iff
\[ \frac{\sin AOB / \sin OAB}{\sin BOC / \sin OCB} = \frac{\sin AOD / \sin OAD}{\sin COD / \sin OCD} \]
But $\angle OAB \equiv \angle OAD$ and $\angle OCB \equiv \angle OCD$ so the equation becomes

\[ \sin AOB \sin COD = \sin AOD \sin BOC \]
depending on only the angles between $OA$, $OB$, $OC$, $OD$.
This is why we can say $O(ABCD)$ is a harmonic pencil.

Suppose $ABC$ is a triangle and the tangents through $A$, $B$ to the circumcircle of $ABC$ intersect at $T$. Then $CT$ is the $C$-symmedian of $ABC$.

Proof: Let the perpendicular bisector of $AB$ (also passing through the center of the circumcircle and through $T$) intersect the circle at $X, Y$, with $Y$ closer to $T$; let the midpoint of $AB$ be $M$.

Then since $M$ and $T$ are inverses w.r.t. the circumcircle, $XMYT$ is a harmonic division. Since $\angle XCY$ is right because $XY$ is a diameter of the circumcircle, $\angle MCT$ is bisected by $CY$. Since $Y$ is the midpoint of arc $AB$, $CY$ also bisects $\angle ACB$. Thus $CM$ and $CT$ are isogonals w.r.t. $\angle ACB$. Since $CM$ is the $C$-median, $CT$ is the $C$-symmedian.
This post has been edited 1 time. Last edited by math_explorer, Aug 19, 2011, 9:26 AM

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