Belarusian Mathematical Olympiad

Let $x = \angle CQI$, $y = \angle QCI$, $2\alpha = \angle BAC$, $2\beta = \angle ABC$, $2\gamma = \angle ACB$. Let $r$ be the inradius of the triangle $ABC$. It is easy to see that

$$\angle CIQ = 360^\circ - 90^\circ - 2\alpha - \gamma =$$
$$= 270^\circ - (\alpha + \beta + \gamma) - \alpha + \beta = 180^\circ - \alpha + \beta.$$
So,
$$x + y = 180^\circ - \angle CIQ = \alpha - \beta. \quad (1)$$
Further,
$$\angle QTL = \angle NTB = 180^\circ - 2(90^\circ - x) = 2x,$$

hence
$$\begin{aligned} \angle ILT &= 0.5\angle ALT = 0.5(2\alpha - 2x) = \alpha - x, \angle INT = 0.5\angle CNT = \\ &= 0.5(2\beta + 2x) = \beta + x. \end{aligned}$$
Then
$$\begin{aligned} LT &= LK - KT = r \cot \angle ILT - r \tan x = r(\cot(\alpha - x) - \tan x), \\ NT &= NK + KT = r \cot \angle INT + r \tan x = r(\cot(\beta + x) + \tan x). \end{aligned}$$
Therefore, from
$$\cot(\alpha - x) - \tan x = \cot(\beta + x) + \tan x \quad (2)$$
it will follow the problem condition.
Indeed, since
$$\begin{aligned} \cot(\alpha - x) - \tan x &= \frac{1 + \tan \alpha \tan x}{\tan \alpha - \tan x} - \tan x = \\ &= \frac{1 + \tan \alpha \tan x - \tan \alpha \tan x + \tan^2 x}{\tan \alpha - \tan x} = \frac{1 + \tan^2 x}{\tan \alpha - \tan x}, \\ \cot(\beta + x) - \tan x &= \frac{1 - \tan \beta \tan x}{\tan \beta + \tan x} + \tan x = \\ &= \frac{1 - \tan \beta \tan x + \tan \beta \tan x + \tan^2 x}{\tan \beta + \tan x} = \frac{1 + \tan^2 x}{\tan \beta + \tan x}, \end{aligned}$$
we see that (2) is equivalent to
$$\tan \alpha - \tan x = \tan \beta + \tan x. \quad (3)$$
By law of sines for $\triangle CQT$, $\frac{r}{\sin y} = \frac{CI}{\sin x}$. Since $CI = r/\sin \gamma$, we have $\sin y = \sin \gamma \cdot \sin x$, i.e. from (1) and the equality $\alpha + \beta + \gamma = 90^\circ$ it follows that $\sin(\alpha - \beta - x) = \sin x \cdot \cos(\alpha + \beta)$. Then
$\sin x \cdot \cos(\alpha + \beta) = \sin(\alpha - \beta) \cos x - \cos(\alpha - \beta) \sin x,$
so $\sin x(\cos(\alpha + \beta) + \cos(\alpha - \beta)) = \sin(\alpha - \beta) \cos x$, which is equivalent to
$2 \sin x \cos \alpha \cos \beta = (\sin \alpha \cos \beta - \cos \alpha \sin \beta) \cos x$, $\iff 2 \tan x = \tan \alpha - \tan \beta$,
i.e. is equivalent to (3), as required.