Q:

Does a right angle triangle ABC, right angled at A has A-symmedian?

A symmedian is defined to be the isogonal of a median in a triangle .
In EGMO , lemma 4.24 (Constructing the Symmedian),which states, "Let $X$ be the intersection of the tangents to $(ABC)$ at $B$ and $C$. Then line $AX$ is a symmedian."
My question is what happens to a right angle triangle , when we do this construction, the tangent lines don't meet . Is this construction of symmedian limited to only acute triangles and obtuse triangles.
The author of the book hasn't commented anything about this .
Though , by a simple angle chase, we can see that for a right angle triangle, the symmedian is the altitude. 
Can someone clarify ?
Note: By EGMO, I mean the book, Euclidean Geometry in Mathematical Olympiads by Evan Chen.

A:

In the projective real plane, if $A$ is a right angle then
the tangents at $B$ and $C$ both are perpendicular to the side $BC$
and therefore are parallel to each other and meet at a point at infinity.
Let this point at infinity be $X$.
Then the line $AX$ is parallel to the tangents at $B$ and $C,$
and therefore it also is perpendicular to side $BC$.
The altitude from $A$ to $BC$ lies along line $AX,$ as required.
You can also do this without a point at infinity if you are willing to consider limiting cases for $A$ an acute angle approaching a right angle and $A$ an obtuse angle approaching a right angle.

