Problem 1526

Define $A^*$, $A^\#$ similarly. Since $A'B' \parallel AB$, $B'C' \parallel BC$, $C'A' \parallel CA$, and the three altitudes are concurrent, we obtain
$$\frac{C'A^\sharp}{A^\sharp B'} \cdot \frac{B'C^\sharp}{C^\sharp A'} \cdot \frac{A'B^\sharp}{B^\sharp C'} = 1 = \frac{BA^*}{A^*C} \cdot \frac{AC^*}{C^*B} \cdot \frac{CB^*}{B^*A} = 1,$$
so $K$ also lies on $A'A^\sharp$. Let $a = BC$, $b = CA$, $c = AB$. By Menelaus' theorem,
$$\frac{A'K}{KA^\sharp} \cdot \frac{A^\sharp C'}{C'B'} \cdot \frac{B'C^\sharp}{C^\sharp A'} = -1 = \frac{A'K}{KA^\sharp} \cdot \frac{A^\sharp A}{AA^*} \cdot \frac{A^*L}{LA'}.$$
Therefore $\frac{A^*L}{LA'} = \frac{AA^*}{A^\sharp A} \cdot \frac{A^\sharp C'}{C'B'} \cdot \frac{B'C^\sharp}{C^\sharp A'} = 2 \cdot \frac{c \cos B}{a} \cdot \frac{b \cos A}{a \cos B} = \frac{2bc \cos A}{a^2}$. Denote this ratio by $r$. And $A'A^* = b \cos C - \frac{1}{2}a = \frac{b \cos C - c \cos B}{2}$, so
$$\begin{aligned} \frac{BL}{LC} &= \frac{c \cos B + \frac{r}{1+r} \cdot \frac{b \cos C - c \cos B}{2}}{b \cos C - \frac{r}{1+r} \cdot \frac{b \cos C - c \cos B}{2}} = \frac{2c \cos B + r(c \cos B + b \cos C)}{2b \cos C + r(c \cos B + b \cos C)} \\ &= \frac{2c \cos B + \frac{2bc \cos A}{a}}{2b \cos C + \frac{2bc \cos A}{a}} \quad (\text{since } c \cos B + b \cos C = a) \\ &= \frac{c(a \cos B + b \cos C)}{b(a \cos C + c \cos A)} = \frac{c^2}{b^2}. \end{aligned}$$
But $\frac{BL}{LC} = \frac{c \sin \angle BAL}{b \sin \angle CAL}$, so we get $\frac{\sin \angle BAL}{\sin \angle CAL} = \frac{c}{b}$. Also, since $A'$ is the midpoint of $BC$, $1 = \frac{BA'}{A'C} = \frac{c \sin \angle BAA'}{b \sin \angle CAA'}$, so
$$\frac{\sin \angle BAL}{\sin \angle CAL} = \frac{\sin \angle CAA'}{\sin \angle BAA'}$$
Because $\angle BAL + \angle CAL = \angle CAA' + \angle BAA' = \angle A$, we conclude $\angle BAL = \angle CAA'$, $\angle CAL = \angle BAA'$. This completes the proof.