# Concurrence Not from School Geometry

### Sohail Farhangi 17 February, 2013

Let the incircle $T$, of $\Delta ABC$ be tangent to the sides at $A'$, $B'$, and $C'$ as shown below. For a point $P$ in the plane, let $A'P$, $B'P$, and $C'P$ intersect $T$ at $A_1$, $B_1$, and $C_1$ respectively. Then $AA_1$, $BB_1$, and $CC_1$ concur at a point $P'$. The applet below serves a dynamic illustration:

### Proof

James Tao recommended to me a proof using the trigonometric form of Ceva's Theorem. First we will use the extended Law of Sines on $\Delta AA_{1}C'$ and $\Delta A_{1}C'A'$ to obtain the following equations where $r$ is the radius of $T$.

 (1) $\displaystyle \frac{C'A_1}{\mbox{sin}(\angle C'AA_{1})}=\frac{AA_1}{\mbox{sin}(\angle AC'A_{1})}$
 (2) $\displaystyle \frac{C'A_1}{\mbox{sin}(\angle C'A'A_{1})}=2r.$

Dividing (2) by (1) gives us the following equation:

 (3) $\displaystyle \frac{\mbox{sin}(\angle C'AA_{1})}{\mbox{sin}(\angle C'A'A_{1})}=\frac{2r\mbox{sin}(\angle AC'A_{1})}{AA_{1}}.$

A similar analysis for $\Delta AB'A_1$ and $\Delta A_{1}1B'A'$ will give us the next equation:
 (4) $\displaystyle \frac{\mbox{sin}(\angle B'AA_{1})}{\mbox{sin}(\angle B'A'A_{1})}=\frac{2r\mbox{sin}(\angle AB'A_{1})}{AA_{1}}.$

Now dividing (3) by (4) yields another equation:

 (5) $\displaystyle \frac{\mbox{sin}(\angle C'AA_{1})\mbox{sin}(\angle B'A'A_{1})}{\mbox{sin}(\angle C'A'A_{1})\mbox{sin}(\angle B'AA_{1})}=\frac{\mbox{sin}(\angle AC'A_{1})}{\mbox{sin}(\angle AB'A_{1})}.$

Now note that $\angle AB'A_{1}=\angle B'A'A_1$ since $AB'$ is tangent to $T$, and similarly we have \angle $AC'A_{1} = \angle C'A'A_1$, which gives us

 (6) $\displaystyle \frac{\mbox{sin}(\angle C'AA_{1})}{\mbox{sin}(\angle B'AA_{1})}=\bigg(\frac{\mbox{sin}(\angle C'A'A_{1})}{\mbox{sin}(\angle B'A'A_{1})}\bigg)^{2}.$

Finally, we are ready to apply the trigonometric form of (the inverse of) Ceva's theorem:

 (7) $\displaystyle \frac{\mbox{sin}(\angle C'AA_{1})\mbox{sin}(\angle B'CC_{1})\mbox{sin}(\angle A'BB_{1})}{\mbox{sin}(\angle A_{1}AB')\mbox{sin}(\angle C_{1}CA')\mbox{sin}(\angle B_{1}BC')}= \\ \displaystyle\bigg(\frac{\mbox{sin}(\angle PA'C')\mbox{sin}(\angle PC'B')\mbox{sin}(\angle PB'A')}{\mbox{sin}(\angle B'A'P)\mbox{sin}(\angle A'C'P)\mbox{sin}(\angle C'B'P)}\bigg)^{2}=1.$ 67609202