Let ABC be a triangle. The tangents of (ABC) from B and C meet at X. AX meet (ABC) at K. Show that BC is the B-symmedian of $\bigtriangleup ABK$ and the C-symmedian of $ \bigtriangleup ACK.$
This problem is from the book Euclidean Geometry in Mathematical Olympiads from Evan Chen (Problem 4.29). I tried to use that ABK is similar to AMC where M is the midpoint of BC. And I also define the isogonal of BD and tried to prove that it is the midpoint. The book says that with the property $\frac{AB}{BK}=\frac{AC}{CK}$ it can be solved.
Let $M$ be the midpoint of $BC$ and let $A'$ be reflection of $A$ with respect to $M$, so $ABA'C$ is a parallelogram and $AA'$ is the $A$ median of $\triangle ABC$.
it is well known that $X$ and $A'$ are isogonal conjugates of each other.
Let $K$ be $AX \cap (ABC) \neq A$.
Let's prove that the tangent lines to $(ABC)$ on $A$ and $K$ meet up on line $BC$: for let $Y$ be the meeting of those two tangents, then line $AK$ is the polar of $Y$ and so we can conclude that $X$ is in the polar of $Y$. By La Hire's theorem, we have that $Y$ is in the polar of $X$ which is the line $BC$. Therefore, $Y,B$ and $C$ are collinear.
We are done, because from my very image which proves that $AX$ is the symmedian of $\triangle ABC$ we must have also that $CY$ is the symmedian of $\triangle ABK$.