Team Selection Test for IMO

Let the incircle of $ABC$ touch the sides $[AB]$ and $[AC]$ at the points $F$ and $E$, respectively. Let $AE=AF=x$, $BD=BF=y$ and $CD=CE=z$. Let the line passing through $D$ and parallel to $AC$ intersect $AB$ at $P$ and the line passing through $D$ and parallel to $AB$ intersect $AC$ at $Q$. Then $BPD\sim BAC$ with similarity ratio $\frac{y}{y+z}$ and $DQC\sim BAC$ with similarity ratio $\frac{z}{y+z}$.

Let $C_1$ and $C_2$ be the incircles of the triangles $BPD$ and $DQC$ with centers $I_1$ and $I_2$, respectively. Let $C_1$ touch the lines $BC$ and $AB$ at the points $U_1$ and $U_2$, respectively, and $C_2$ touch the lines $BC$ and $AC$ at the points $V_1$ and $V_2$, respectively. By the similarity we have $D{U}_1=\frac{yz}{y+z}$, $B{U}_2=\frac{y^2}{y+z}$, $D{V}_1=\frac{yz}{y+z}$ and $C{V}_2=\frac{z^2}{y+z}$. As $D{U}_1=D{V}_1$, we see that $D$ is on the radical axis of $C_1$ and $C_2$.

$A{U}_2=AB-B{U}_2=x+y-\frac{y^2}{y+z}=\frac{xy+yz+xz}{y+z}$ and $A{V}_2=AC-C{V}_2=x+z-\frac{z^2}{y+z}=\frac{xy+yz+xz}{y+z}$. Therefore $A$ is on the radical axis of $C_1$ and $C_2$ as well and hence $AC$ is the radical axis of these circles. Thus we have $AD\perp I_1{I}_2$.

Note that the intersection of the lines $B{I}_1$ and $C{I}_2$ is $I$. That is easy to see that $D{I}_1{I}_2$ is a parallelogram. Therefore $T$ is the intersection point of the line segments $[ID]$ and $[I_1{I}_2]$, hence $M,I_1,I_2$ and $N$ are collinear. As $P{U}_1\parallel AD$, we have $I_{1}M\perp P{U}_1$. The similarity of $BPD$ and $BAC$ gives $\frac{BK}{BM}=\frac{y+z}{y}$, that is $\frac{KM}{BM}=\frac{z}{y}$. Similarly we get $\frac{LN}{CN}=\frac{y}{z}$ and we are done.