Mongolian National Mathematical Olympiad

Let $\omega$ be the incircle of the triangle $ABC$, $A_1$, $B_1$, $C_1$ be the touching points of $\omega$ with sides $BC$, $AC$ and $AB$ respectively. Let ${\omega }^{\prime }$ be the circle with diameter $EI$, and $K$ be the intersection point of $\omega$ and ${\omega }^{\prime }$, different from $A_1$. Denote by $M$ be the intersection point of $AK$ and ${\omega }^{\prime }$. We claim that $I,M$ and $D$ are collinear. For this, it suffices to show that $E,C,D^{\prime }$ and $B$ are harmonic because $D$ is uniquely determined by the given ratio, where $D^{\prime }$ is the intersection point of $IM$ and $BC$. Since $\angle ML{A}_1=\angle K{C}_1{A}_1=\angle KIE=\angle KME$ we get $L{A}_1\parallel EM$ and so $L{A}_1\perp MI$, i.e, $D^{\prime }L$ is tangent to the circle $\omega$. Thus $$sin(\angle D^{\prime }IB)=\sin (\angle LI{C}_1/2)\text{ and }\sin (\angle CI{D}^{\prime })=\sin (\angle B_{1}IL/2).(\ast )$$ Clearly, the points $K,B_1,L,C_1$ are harmonic and the cross-ratio is defined as a ratio of sines. Therefore we can conclude that $E,C,D^{\prime }$ and $B$ are harmonic since $\sin (\angle EIB)=\sin (\angle KI{C}_1/2)$, $\sin (\angle EIC)=\sin (\angle KI{B}_1/2)$ and $(\ast )$. Hence $D=D^{\prime }$, implying the claim. Let $H^{\prime }$ be the intersection point of $EI$ and $AK$. Since $\angle KEI=\angle KMI=\angle IE{A}_1$ the points $H^{\prime },E,M$ and $D$ lie on a circle. Thus $H=H^{\prime }$ and $\angle A{H}^{\prime }E=\angle I{D}^{\prime }E$.