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Geometry Difficulty 6.3 National Olympiad Prove it Vietnam

In the plane fix two points $A, B$ ( $A \neq B$ ). Consider a point $C$ moving in the plane such that $\overrightarrow{ACB} = \alpha$ , where $\alpha$ is a given angle ( $0^\circ < \alpha < 180^\circ$ ). The circle with center in $I$ , inscribed the triangle $ABC$ touches the sides $AB$ , $BC$ and $CA$ in the points $D$ , $E$ and $F$ , respectively. The line $AI$ and $BI$ intersect $EF$ in $M$ and $N$ , respectively. Show that:
1/ The segment $MN$ has constant length;
2/ The circumcircle of the triangle $DMN$ goes through a fix-point.

Solution

1/ Consider the triangle $AFM$ , we have:
$\begin{align*} \widehat{AMF} &= 180^\circ - (\widehat{MFA} + \widehat{FAM}) = 90^\circ - (\widehat{EFI} + \widehat{FAM}) = 90^\circ - (\widehat{ECI} + \widehat{FAM}) \\ &= 90^\circ - \left(\frac{C}{2} + \frac{A}{2}\right) = \frac{B}{2} = \widehat{IBA}. \end{align*}$
We also have: $\widehat{NIM} = \widehat{AIB}$ .
Hence: $\triangle IMN \sim \triangle IBA$ . (*)
Hence, draw $IH \perp MN$ , we obtain:
$\frac{MN}{BA} = \frac{IH}{ID} = \frac{IH}{IF} = \sin \widehat{EFI} = \sin \frac{\alpha}{2}.$
Consequently: $MN = BA \cdot \sin \frac{\alpha}{2} = \text{const.}$

2/ It is easy to see that the points $F$ and $D$ are symmetric with respect to the line $AM$ . According to (*), we have: $\widehat{IMD} = \widehat{IBD}$ . Thus, the quadrilateral $\widehat{IMBD}$ is cyclic.
Hence $\widehat{BMA} = 90^\circ$ , in other words $\widehat{BMA}$ is a right triangle in $\widehat{M}$ . Hence, let $P$ be the middle point of $AB$ , we have $\widehat{BPM} = 2\widehat{BAM} = \widehat{BAC}$ .
Since the point $E$ and $D$ are symmetric with respect to the line $BN$ , according to (*)
we have:
$\widehat{MND} = 2\widehat{INM} = 2\widehat{IAB} = \widehat{BAC}.$
Thus, we have $\widehat{BPM} = \widehat{MND}$ . Consequently the points $M, N, D, P$ lie in one circle.
This means the circumcircle of $\triangle DMN$ goes through the fix point $P$ - the middle point of $AB$ .

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