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I read about the fallacy of isosceles triangle

Incorrect figure, enter image description here Correct figure, enter image description here 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)

CandidFlakes
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  • You talk of an isosceles triangle, but none of the two diagrams seems to be of an isosceles triangle...and the link is dead. And the claim cannot be true in the general case as an equilateral triangle is also an isosceles one, and there the angle bisector of the top vertex and the perpendicular bisector of the base are one and the same... – DonAntonio Nov 19 '16 at 09:08
  • You can visit the link-https://brilliant.org/discussions/thread/fallacy-of-the-isosceles-triangle/ – CandidFlakes Nov 19 '16 at 09:13

2 Answers2

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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 $\triangle(ABC)$.

enter image description here

Christian Blatter
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  • Isn't it possible for AM to intersect the perpendicular bisector of BC inside the triangle or on the triangle?how con you be sure that the bisector of angle(A) intersects the perpendicular bisector of BC at M? – CandidFlakes Nov 19 '16 at 12:51
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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.

DonAntonio
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