I conjecture that $MACB$ is a cyclic quadrilateral, but I can't finish the problem with this fact alone. Interesting problem!

hsiangshen wrote: $\triangle ABC$, $\angle A=23^{\circ},\angle B=46^{\circ}$. Let $\Gamma$ be a circle with center $C$, radius $AC$. Let the external angle bisector of $\angle B$ intersects $\Gamma$ at $M,N$. Find $\angle MAN$. Angle chase/

Abhaysingh2003 wrote: hsiangshen wrote: $\triangle ABC$, $\angle A=23^{\circ},\angle B=46^{\circ}$. Let $\Gamma$ be a circle with center $C$, radius $AC$. Let the external angle bisector of $\angle B$ intersects $\Gamma$ at $M,N$. Find $\angle MAN$. Angle chase/ Thanks for the extremely enlightening and insightful solution. Indeed a wonderful addition to this post

Let $C_{\perp}$ be the projection of $C$ on the internal angle bisector of $\angle B$. Let $C'$ be the projection of $C$ on $MN$. Since $CC_{\perp}BC'$ is a rectangle (internal bisector $\perp$ external bisector), we have $CC' = BC_{\perp} = BC\cos(23 ^{\circ})$. By Pythagorean, we have $2R\sin(\angle MAN) = MN = 2MC' = 2\sqrt{AC^2-BC^2\cos^2(23^{\circ})}$ Simplifying, we have $\cos^2(\angle MAN) = \frac{BC^2}{AC^2} \cdot \cos^2(23^{\circ})$ Notice by LOS, we have $\frac{BC}{AC} = \frac{\sin(23^{\circ})}{\sin(46^{\circ})} = \frac{1}{2\cos(23^{\circ})}$ Thus, we have $\cos^2(\angle MAN) = \frac{1}{4}$, and since $A$ is on the same side of $MN$ as $C$ (by definition, of external angle bisectors when $\angle ABC < 180^{\circ}$), we have $\angle MAN = 60^{\circ}$

Inconsistent wrote: Let $C_{\perp}$ be the projection of $C$ on the internal angle bisector of $\angle B$. Let $C'$ be the projection of $C$ on $MN$. Since $CC_{\perp}BC'$ is a rectangle (internal bisector $\perp$ external bisector), we have $CC' = BC_{\perp} = BC\cos(23 ^{\circ})$. By Pythagorean, we have $2R\sin(\angle MAN) = MN = 2MC' = 2\sqrt{AC^2-BC^2\cos^2(23^{\circ})}$ Simplifying, we have $\cos^2(\angle MAN) = \frac{BC^2}{AC^2} \cdot \cos^2(23^{\circ})$ Notice by LOS, we have $\frac{BC}{AC} = \frac{\sin(23^{\circ})}{\sin(46^{\circ})} = \frac{1}{2\cos(23^{\circ})}$ Thus, we have $\cos^2(\angle MAN) = \frac{1}{4}$, and since $A$ is on the same side of $MN$ as $C$ (by definition, of external angle bisectors when $\angle ABC < 180^{\circ}$), we have $\angle MAN = 60^{\circ}$ Nice solution! What led you to consider the rectangle?

Nice sol! Looking forward to seeing pure geo sol Most of my friends uses trig. too but I think there's a nice pure geo solution...

potatothegeek wrote: Nice solution! What led you to consider the rectangle? We know the angle $\angle MAN$ is intrinsically tied to the length of $MN$, which is tied to the distance of that line from $C$, which is tied through trig to these projections. The rectangle was just there to make the trig ever so easier.

potatothegeek wrote: Abhaysingh2003 wrote: hsiangshen wrote: $\triangle ABC$, $\angle A=23^{\circ},\angle B=46^{\circ}$. Let $\Gamma$ be a circle with center $C$, radius $AC$. Let the external angle bisector of $\angle B$ intersects $\Gamma$ at $M,N$. Find $\angle MAN$. Angle chase/ Thanks for the extremely enlightening and insightful solution. Indeed a wonderful addition to this post You are welcome my dear!

Ok so here's a solution from my friend Hakurei_Reimu Let $AB$ intersects $\Gamma$ at $P$ We can easily see PERPENDICULAR! (And can be easily proved by angle chasing) The rest is easy

Attachments: