Problem 2266

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 $DU_1 = \frac{yz}{y+z}$, $BU_2 = \frac{y^2}{y+z}$, $DV_1 = \frac{yz}{y+z}$ and $CV_2 = \frac{z^2}{y+z}$. As $DU_1 = DV_1$, we see that $D$ is on the radical axis of $C_1$ and $C_2$.

$AU_2 = AB - BU_2 = x + y - \frac{y^2}{y+z} = \frac{xy + yz + xz}{y+z}$ and $AV_2 = AC - CV_2 = x + z - \frac{z^2}{y+z} = \frac{xy + yz + xz}{y+z}$. 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 $AD \perp I_1I_2$.

Note that the intersection of the lines $BI_1$ and $CI_2$ is $I$. That is easy to see that $DI_1I_2$ is a parallelogram. Therefore $T$ is the intersection point of the line segments $[ID]$ and $[I_1I_2]$, hence $M, I_1, I_2$ and $N$ are collinear. As $PU_1 \parallel AD$, we have $I_1M \perp PU_1$. 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.