How do we come to know that they intersect each other outside the triangle?

Question

I read about the fallacy of isosceles triangle

Incorrect figure, Correct figure, The description states that the point of intersection of the bisector of angle(A) and the perpendicular bisector of side BC intersect each other at point D outside the triangle and not inside the triangle. But,my question is that how do we come to know that they intersect each other outside the triangle and not inside the triangle? Can you prove that they intersect each other outside the triangle? Or, Do we need to construct accurate figure to come to know that they intersect each other outside the triangle ?(but,most of the time we do not need to construct figure as only the rough figure works as for example "prove that the opposite sides of a parallelogram are equal",here,we need not to construct a parallelogram)

Answer 1

Now the link works, and the question "how do we come to know they instersect outside the triangle and not inside?" has the answer that: otherwise, you have proved the triangle must be isosceles (in fact, equilateral but we don't care much for that now) and, thus, we know the top vertex's angle bisector is also the height to the opposite side (the triangle's base) and thus it is also the base's perpendicular bisector, and thus of course there is no " one point intersection $\;D\;$" as shown, i.e.: contradiction.

That's why we know both lines must intersect outside the triangle, as all the proof's steps are correct except the very last one, when we assume $\;\;$ is inside the triangle.

Answer 2

Draw the circumcircle of $\triangle(ABC)$. The angle bisector at $A$ intersects this circle at $A$ and (by the inscribed angle theorem) at the point $M$ halving the lower arc $BC$ of this circle. This point $M$ is also on the perpendicular bisector of $BC$, hence the point of intersection $P$ of the two lines in question. It follows that $P$ is lying outside $\triangle(ABC)$.