My solution(Same as official solution): Let $BK \cap CL=T$. Then $AT$ is a symmedian, and so $AT$ is isogonal to $AG$, where $G$ is the centroid of $\triangle ABC$. Then the result follows by Isogonal Line Lemma on $AG$ and $AT$.

Kinda unbelievable that this was the third problem of day 2. Line through $BK\cap CL$ and $A$ is the symmedian and the line through $A$ and $BL\cap CK$ are median. So by isogonal line lemma, the required angles are equal.

26 people solved it on the exam

if $BAK=x$ and $CAL=y$ after 4 sines law in triangles $CAL ,BCL ,BCK ,BAK$ and multiplying relations we've got we'll arrive : $\dfrac{sinA+x}{sinx} =\dfrac{sinA+y}{siny}$ multiply it by $-1$ we'll arrive this famous lemma : let $x ,y ,z ,t$ be for angles such that $\dfrac{sinx}{siny}=\dfrac{sinz}{sint}$ and $x+y =z+t$ then we have $x=z$ and $y=t$ and we're done.

Lets use barycentric coordinates. A= (1:0:0), B=(0:1:0) , C=(0:0:1). Then calculating the points in the tangent line B is in the form (x : 1 : -x ) . And the points in $CC_1$ have y=x. So the point K is (-1:1:1) and L is (1:1:-1). One is isogonal conjugate of the another because is (1ˆ2 / -1: 1ˆ2 /1 : 1ˆ2/1). Finished.