I'm a high school student and I had this theorem in my book:-
The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing angle.
Now this theorem confused me. Why? because I used it on an equilateral triangle and it didn't work (the angle bisector and opposite line extended were parallel). Hence I searched it on google, finding wikipedia's article:- Wikipedia's page of angle bisector theorem
It mentions that the theorem applies to non-equilateral triangle. Now I create some other triangles (one of angles 90, 45, 45 and other of 80, 50, 50).
Now we can see that in both the figures above $\measuredangle B = bisector\space of \measuredangle A$
Hence it can be shown that $bisector\space of \measuredangle A\space ||\space BC$ so they'll never intersect each other.
In that case, this theorem can't be applied. Why then, it is a theorem if there are so many contradictions to this, or I'm getting something wrong?
Edit:- The answer by SotArmen made me think another thing.
The theorem says that in the figures like the above, $\frac {AB}{AC} = \frac {BD}{CD}$
here, as the triangles are isosceles, $\frac {AB}{AC} = 1$ and hence $\frac {BD}{CD} = 1$
Now, $BD$ and $CD$ are of same length, but we know that $C$ cannot be to the left of $B$ neither can it be to the right of $D$. So there is only one possibility here, that $B$ and $C$ coincide, but then everything breaks down..., it's no longer a triangle. So, the theorem is wrong in this case I guess?
It actually still holds, but in a more weird way.
Think if the parallel line hit the oposite side of the triangle, eg in your first figure on point D and let $BC=x$ then the angle bisector theorem says that:
$$ \frac{BD}{BD +x} =1 \implies \frac{x}{BD+x}=0 $$
Okay, so far this is obviously a contradiction, but think what happens whenever $BD$ becomes huge, this goes close to zero.
Actually , we normally think for those kinds of theorems that there is a point in infinity, one may think of it as the horizon such that all lines meet there, infinitely far away.
Now, if BD was infinite in the above the fraction would be zero and the theorem would hold right?
This is one of the justifications of the theorem in this case.
Add on after you last edit:
The point is that if the point of intersection is infinitely far away, the distances are in essense, equal.
This can be done rigorously, you cqn search for stereographic projection, but I will try to give a more intuitive explanation nonetheless.
Think of a very large corridor, if you stare directly, it looks as if there is a point in the middle where the walls meet.
Now think as if you are inside the triangle and stare at the infinitely long parallel lines BC and the bisector then it would seem as they meet somewhere in the horizon. In that point in the horizon the distance to C and to B is the same, they are both infinite and the theorem "holds".
This play with perspective is used heavily in art to give a more aesthetically pleasing result , which is a very cute connection between art and geometry.