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 $\omega'$ be the circle with diameter $EI$, and $K$ be the intersection point of $\omega$ and $\omega'$, different from $A_1$. Denote by $M$ be the intersection point of $AK$ and $\omega'$. We claim that $I, M$ and $D$ are collinear. For this, it suffices to show that $E, C, D'$ and $B$ are harmonic because $D$ is uniquely determined by the given ratio, where $D'$ is the intersection point of $IM$ and $BC$. Since $\angle MLA_1 = \angle KC_1A_1 = \angle KIE = \angle KME$ we get $LA_1 \parallel EM$ and so $LA_1 \perp MI$, i.e, $D'L$ is tangent to the circle $\omega$. Thus
$$sin(\angle D'IB) = \sin(\angle LIC_1/2) \text{ and } \sin(\angle CID') = \sin(\angle B_1IL/2). \quad (*)$$
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 $E, C, D'$ and $B$ are harmonic since $\sin(\angle EIB) = \sin(\angle KIC_1/2)$, $\sin(\angle EIC) = \sin(\angle KIB_1/2)$ and $(*)$. Hence $D = D'$, implying the claim.
Let $H'$ be the intersection point of $EI$ and $AK$. Since $\angle KEI = \angle KMI = \angle IEA_1$ the points $H', E, M$ and $D$ lie on a circle. Thus $H = H'$ and $\angle AH'E = \angle ID'E$.