Invert at $A$ with arbitrary radius to obtain the following equivalent problem: Inverted Problem wrote: In $\triangle ABC$, let $P$ and $Q$ be the feet of the interior and exterior angle bisectors of $\angle BAC$ on $\overline{BC}$, respectively. Let the line through $Q$ perpendicular to $\overline{QA}$ intersect $\overline{AB}$ at $M$, and let $N=\overline{AP} \cap \overline{MC}$. Prove that $P$ is the midpoint of $AN$. This is simply \[ -1 = (QP; BC) \stackrel{A}{=} (\overline{AQ} \cap \overline{MC}, N; MC) \stackrel{Q}{=} (AN; \overline{BN}_{\infty}, P), \]as desired.

Let $X$, $Y$, and $Z$ be the midpoints of sides $AP$, $BC$, and $MC$, respectively. It is obvious that $\angle APB = \angle CAB$. Since $QMYC$ is cyclic, we have $\angle QYM = \angle QCM = \angle QCA = \angle QPA$. This implies that $MY \parallel AP$. Therefore, $\angle YMC = \angle PAC = \angle BAX$. It is easy to see that $\triangle BAP \sim \triangle YMC$, and consequently, $\triangle BAX \sim \triangle YMZ$. Thus, $\angle ABX + \angle BAX = \angle MYZ + \angle YMZ = \angle BMY + \angle YMZ = \angle BMZ$. The problem follows immediately. $\blacksquare$

ppanther wrote: Invert at $A$ with arbitrary radius to obtain the following equivalent problem: Inverted Problem wrote: In $\triangle ABC$, let $P$ and $Q$ be the feet of the interior and exterior angle bisectors of $\angle BAC$ on $\overline{BC}$, respectively. Let the line through $Q$ perpendicular to $\overline{QA}$ intersect $\overline{AB}$ at $M$, and let $N=\overline{AP} \cap \overline{MC}$. Prove that $P$ is the midpoint of $AN$. This is simply \[ -1 = (QP; BC) \stackrel{A}{=} (\overline{AQ} \cap \overline{MC}, N; MC) \stackrel{Q}{=} (AN; \overline{BN}_{\infty}, P), \]as desired. Can you please explain your solution??

So my solution goes as follows. Let $K$ be the midpoint of $BC$. Let $R$ be the intersection of $(AMB)$ and $AP$. I claim that $AR=RP$. Note that $\triangle MRB\sim\triangle CPB$, since $\angle RBM=\angle BMR=\angle BAP=\angle BCP=\angle PBC$. By the spiral similarity, $\triangle RPB\sim\triangle MCB$, hence $\dfrac{RP}{MC}=\dfrac{BP}{BC}$. We have $\triangle QMC\sim \triangle QAP$, since $\angle QMC=\angle QAP=90^{\circ}$ and $\angle MCQ=\angle APQ$, thus $\dfrac{AQ}{QM}=\dfrac{AP}{CM}$. Also, $\triangle AQM\sim \triangle PBK$, since $\angle BKP=\angle AMQ=90^{\circ}$ and $\angle MAQ=\angle KPB$, thus $\dfrac{BP}{BK}=\dfrac{AQ}{AM}$. \begin{align*} RP&=\frac{MC\cdot BP}{BC}\\ &=\frac{MC\cdot BP}{2\cdot BK}\\ &= \frac{1}{2}\cdot \frac{MC\cdot AQ}{QM} \\ &=\frac{1}{2}\cdot AP \end{align*}Therefore, $AR=RP$, we are done.

This was literally nothing with complex numbers...