, 2013

Because $LM$ is parallel to $BC$, the problem is equivalent to proving that $\angle AA_{1}L = \angle AML = \angle ACB$. We present two solutions:

First solution. Let $L_{1}$ be the point on $AB$ such that $\angle AA_{1}L_{1} = \angle ACB$ and let us prove that $L_{1} = L$.

Quadrilateral $BCB_{1}C_{1}$ is cyclic since $\angle BC_{1}C = \angle BB_{1}C = 90^{\circ}$. Therefore, $\angle B_{1}C_{1}A = \angle ACB = \angle AA_{1}L_{1}$. We deduce that the quadrilateral $A_{1}KC_{1}L_{1}$ is cyclic. Hence, $\angle L_{1}KA_{1} = \angle L_{1}C_{1}A_{1}$.

Quadrilateral $AC_{1}A_{1}C$ is cyclic since $\angle CA_{1}A = \angle CC_{1}A = 90^{\circ}$. We deduce that $\angle L_{1}C_{1}A_{1} = \angle ACB$. Thus, $\angle L_{1}KA_{1} = \angle AA_{1}L_{1}$. This proves that $L_{1}A_{1} = L_{1}K$, that is $L_{1} = L$, the intersection point of the perpendicular bisector of $KA_{1}$ with $AB$.

Second solution. Since $LK = LA_{1}$, the problem is equivalent to proving that $\angle LKA_{1} = \angle ACB$. But $\angle BHA_{1} = 90^{\circ} - \angle A_{1}BH = \angle ACB$, where $H$ is the orthocenter of triangle $ABC$. Therefore, the problem is equivalent to proving that $LK$ and $BH$ are parallel.

We know that $\frac{AL}{LB} = \frac{AN}{NA_{1}}$, where $N$ is the midpoint of $KA_{1}$, since $LM$ and $BC$ are parallel. It remains to prove that $\frac{AK}{KH} = \frac{AN}{NA_{1}}$.

To simplify the notations let us write $\alpha = \angle BAC$, $\beta = \angle CBA$ and $\gamma = \angle ACB$.

We know that $\angle KC_{1}A = \gamma$, $\angle HC_{1}K = 90^{\circ} - \gamma$, $\angle C_{1}AK = 90^{\circ} - \beta$ and $\angle KHC_{1} = \beta$. We deduce by applying sine laws on triangles $AC_{1}K$ and $C_{1}HK$ that
$$\frac{AK}{KH} = \frac{\sin \angle KC_{1}A \cdot \sin \angle KHC_{1}}{\sin \angle HC_{1}K \cdot \sin \angle C_{1}AK} = \tan \beta \cdot \tan \gamma$$
On the other hand, we have $\frac{AN}{NA_{1}} = \frac{AK}{NA_{1}} + 1 = 2 \frac{AK}{KA_{1}} + 1$. We also know that $\angle A_{1}C_{1}K = 180^{\circ} - 2\gamma$ and $\angle KA_{1}C_{1} = 90^{\circ} - \alpha$. We deduce by applying sine laws on triangles $AC_{1}K$ and $C_{1}A_{1}K$ that
$$\frac{AK}{KA_{1}} = \frac{\sin \gamma \cdot \cos \alpha}{\sin 2\gamma \cdot \cos \beta} = \frac{\cos \alpha}{2 \cos \beta \cdot \cos \gamma}$$
We deduce that
$$\frac{AN}{NA_{1}} = \frac{\cos \alpha}{\cos \beta \cdot \cos \gamma} + 1 = \frac{\cos \beta \cdot \cos \gamma - \cos (\beta + \gamma)}{\cos \beta \cdot \cos \gamma} = \tan \beta \cdot \tan \gamma = \frac{AK}{KH}.$$