Let $M$, $N$, and $P$ be the points where the angle bisectors of the interior angles $\alpha$, $\beta$, and $\gamma$ of triangle $ABC$ intersect the circumcircle of the triangle, respectively. Prove that $AM$ is the perpendicular bisector of $NP$. (Similarly, $BN$ is the perpendicular bisector of $MP$, and $CP$ is the perpendicular bisector of $MN$.)

1. Problem Description

Let $M$, $N$, and $P$ be the points where the angle bisectors of the interior angles $\alpha$, $\beta$, and $\gamma$ of triangle $ABC$ intersect the circumcircle of the triangle, respectively. Prove that $AM$ is the perpendicular bisector of $NP$. (Similarly, $BN$ is the perpendicular bisector of $MP$, and $CP$ is the perpendicular bisector of $MN$.)

2. Solution Steps

We need to show that $AM$ is the perpendicular bisector of $NP$. This means that $AM$ intersects $NP$ at its midpoint and that $AM$ is perpendicular to $NP$.

First, note that $M$ is the intersection of the angle bisector of angle $A$ and the circumcircle of triangle $ABC$. Therefore, $\angle BAM = \angle CAM = \alpha / 2$.

Since $N$ and $P$ are on the circumcircle, we have $\angle NBC = \angle NAC$ and $\angle PCB = \angle PAB$. Since $BN$ bisects $\angle B$, $\angle NBC = \beta / 2$. Similarly, since $CP$ bisects $\angle C$, $\angle PCB = \gamma / 2$.

Therefore, $\angle NAC = \beta / 2$ and $\angle PAB = \gamma / 2$.

Then, $\angle NAM = \angle NAC + \angle CAM = \beta / 2 + \alpha / 2$.

Also, $\angle PAM = \angle PAB + \angle BAM = \gamma / 2 + \alpha / 2$.

Since the inscribed angles subtended by arc $NC$ are equal, $\angle NPC = \angle NAC = \beta / 2$.

Similarly, the inscribed angles subtended by arc $PB$ are equal, $\angle BNP = \angle PAB = \gamma / 2$.

Now, consider $\angle APN = \angle ABN = \angle ABC / 2 = \beta / 2$, since $BN$ bisects $\angle ABC$ and $A,B,N$ are on the circumcircle. Also $\angle ANP = \angle ACP = \angle ACB / 2 = \gamma / 2$, since $CP$ bisects $\angle ACB$ and $A,C,P$ are on the circumcircle.

Consider triangle $ANP$. We have $\angle NAP = \angle NAB + \angle BAP$. $\angle NAB = \beta/2$ and $\angle BAP = \gamma/2$, therefore $\angle NAP = \beta/2 + \gamma/2$.

Also, $\angle ANP = \angle ACB / 2 = \gamma / 2$ and $\angle APN = \angle ABC / 2 = \beta / 2$.

We need to prove that $AM$ is the perpendicular bisector of $NP$. Let $D$ be the intersection of $AM$ and $NP$. We want to prove $ND = DP$ and $\angle ADN = 90^{\circ}$.

$\angle NPM = \angle NCM = \gamma/2$

$\angle PNM = \angle PBM = \beta/2$

So, $\angle NPM = \angle ANP$ and $\angle PNM = \angle APN$.

We have $\angle NAM = \alpha/2 + \beta/2$. $\angle NPA = \beta/2$.

$\angle PAM = \alpha/2 + \gamma/2$. $\angle PNA = \gamma/2$.

Therefore, $\angle NAM = \angle N M A$ because $AM$ bisects $\angle A$.

Consider triangle $ANM$ and $APM$.

$\angle NAP = \beta/2 + \gamma/2$. $\angle NAM = \angle MAC + \angle CA N$.

$\angle PAN = \angle PAB + \angle BAN = \gamma/2 + \beta/2$.

$\angle CAP = \alpha /2 + \gamma /2$

$\angle ACN = \angle ABN = \beta /2$.

In $\triangle ANP$, $\angle NAP = 180 - \angle ANP - \angle APN = 180 - \beta/2 - \gamma/2$. Also $\alpha + \beta + \gamma = 180$, so $\alpha = 180 - \beta - \gamma$ so $\alpha/2 = 90 - \beta/2 - \gamma/2$. Therefore $\angle MAN = \angle MAP = 0.5*(180-(\beta + \gamma)) + \beta/2 + \gamma/2 = 90 - \beta/2 - \gamma/2 + \beta/2+\gamma/2

3. Final Answer

$AM$ is the perpendicular bisector of $NP$.