An equilateral triangle and a 60° angle drawn from the circumcircle form two angle bisectors.

Question

I do not have the original source of this problem, but it appears (with a sligthly different statement) in the book Solving problems in geometry by Hang, Wang.

Let $\triangle ABC$ be equilateral and $D$ the antipode of $A$ in its circumcircle. Consider points $M$ in $AB$ and $N$ in $AC$ such that $\angle MDN=60^\circ$. Show that $\overleftrightarrow{DM}$ is the angle bisector of $\angle BMN$ and $\overleftrightarrow{DN}$ is the angle bisector of $\angle MNC$.

The only solution I now feels a bit unnatural too me and I would like to see a different one if possible. I tried for a while to prove that the projection from $D$ to $MN$ formed two congruent triangles, so when I saw this other solution I got disapointed.

The one I know goes as follows:

Consider $X$ in $\overleftrightarrow{AC}$ such that $\overline{BM}=\overline{CX}$. Then $\triangle MBD\cong \triangle DCX$ and then $\triangle DMN\cong \triangle DXC$. Hence $\angle BMD=\angle DXC=\angle DXN=\angle DMN$.

Answer 1

Last statement of book solution is trying to make use of cyclic quadrilateral without clearly proving there is one. Following is my solution.

First note that $BD=CD$ and $\angle BDM+ \angle CDN=60^\circ$.

Reflect $B$ in $MD$ to get $B'$. Then $\triangle BDM \cong \triangle B'DM$.

Similarly reflect $C$ in $CN$ to get $C'$. So $\triangle CDN \cong \triangle C'DN$.

But $B'D=BD=CD=C'D$ and $\angle MDB' + \angle NDC' = \angle MDB + \angle NDC = 60^\circ = \angle MDN$.

This means $B'D$ and $C'D$ coincide and $B'$ and $C'$ are same points.

Finally $\angle MB'D + \angle NB'D = \angle MBD + \angle NBD = 90^\circ + 90^\circ = 180^\circ$. This shows $B'$ lies on $MN$.

Thus $B'$ is projection of $D$ on $MN$ (as you wanted to) and $DM$ and $DN$ can be observed as angle bisectors of $\angle BMN$ and $\angle MNC$ respectively.

Answer 2

Here is another solution using series of cyclic quadrilaterals and angle chasing.

First note that segment $MY$ subtends $60^\circ$ at both points $B$ and $D$, and as both points $B$ and $D$ are on the same side of the segment $MY$, $BDYM$ is cyclic. Similary, segment $XD$ subtends $30^\circ$ at both points $C$ and $D$ and we conclude $XNCD$ is cyclic.

So, $\angle BMD = \angle BYD = \angle BCD + \theta = 30^\circ + \theta$

$\angle DMY = \angle DBY = 30^\circ$, $\angle DNX = \angle DCX = 30^\circ$.

As $\angle XMY = \angle XNY = 30^\circ$. $MNYX$ is cyclic.

So, $\angle YMN = \angle YXN = \angle NDC = \theta$.

That shows $\angle BMD = \angle NMD = 30^\circ + \theta$ and $MD$ is angle bisector of $\angle BMN$.

Now, $\angle DNC = \angle DXC = \angle BDM + \angle DBC = \angle BDM + 30^\circ$

As $\angle BDM = \angle BYM = \angle MNX$, we have $\angle DNC = \angle DNM$. So $ND$ is angle bisector of $\angle MNC$.