I'm attempting to prove what is demonstrated in this Wolfram demo, http://demonstrations.wolfram.com/TheIntersectionOfAnAngleBisectorAndAPerpendicularBisector/: That the angle bisector of angle A in triangle ABC intersects the perpendicular bisector of BC at a point on the circumcircle of ABC, outside ABC. My professor mentioned that it was always true but that we didn't have time in class for a proof, and I'm very curious about how it can be proved. However, I'm not really sure where to start. I attempted to assign coordinates to the relevant lines, but that quickly got very messy (possibly because I picked the wrong origin). How would I go about doing this proof? Any pointers would be very appreciated. Thanks!
[Math] Intersection of angle bisector and perpendicular bisector of opposite side
angleeuclidean-geometrygeometrytriangles
Related Solutions
Assume the given triangle $ABC$ is acute. Draw the circle with center $M$ through $B$, $C$, $B'$, $C'$; then draw tangents from $A$ (which lies outside) to the circle. Draw a red line $g$ through the two tangent points; this line intersects the line $B\vee C$ in a point $D$.
Our drawing plane can be embedded in a projective plane $P$ by adding elements at infinity (we won't need them). At any rate there is a perspective involution $\iota:\ P\to P$ with center $A$ which keeps $g$ pointwise fixed, and which has the following additional property: For any line $\ell$ through $A$ that intersects the circle it interchanges the two intersection points. (Projectively, this $\iota$ is conjugate to an ordinary reflection $\iota'$ in the $x$-axis $g'$. The map $\iota'$ has its center $A'$ at $y=\pm\infty$, keeps $g'$ pointwise fixed and interchanges points on the unit circle that lie on the same vertical.)
In particular $\iota(B)=C'$, $\iota(C')=B$, $\iota(C)=B'$, $\iota(B')=C$. Therefore $\iota(C'\vee B')=B\vee C$. Therefore $C'\vee B'$ and $B\vee C$ intersect on $g$, which implies that our point $D$ is the same as the point $D$ in the question. In the same way it follows that $B\vee B'$ and $C\vee C'$ intersect on $g$, which implies that the point $H$ lies on $g$ as well.
Since the line $A\vee M$ is orthogonal to $g$, it is orthogonal to $D\vee H$, as stated.
First off, the other two answers here are wrong because you can prove the statement using the same flaw if P was outside the triangle. The proof using the same flaw but done in the outside of the triangle is here: [https://www.youtube.com/watch?v=Yajonhixy4g][1]
The video clearly does this proof with P on the outside. If P were on the segments of the triangle, then the proof will still hold because triangle AEP will be congruent to triangle AFP by SAA (shared sides, bisected angles, 90 degree angles). You know that P and D are the same points because the definition of P is where the perpendicular bisector meets the bisector of the opposite angle, and provided the case where P's on BC, it can only possibly be a triangle if P and D are on the same place. D is the bisector of BC, so so is P. EP = FP because we proved that AEP = AFP. BP = CP because P bisects and is between BC. Since EBP and FCP are right triangles and the Hypotenuse and one of the legs are congruent, we have triangle EBP is congruent to triangle FCP. From here, basic elementary algebra can be use to prove AB = AC.
So, what is wrong with this proof? The answer is betweenness. Every single one of these proofs rely on a certain order of points your drawing above should be:
We still have AE=AF, PE=PF, and PB=PC, and it's still true that BE=FC, but AB=AC is not true. This is because F is still between A and C, but E is not between A and B. Since E is not between A and B, SAA does not follow, and neither does SAS for this particular case.
Best Answer
It is easier to prove the following equivalent version:
Proof: since $\widehat{BAN}=\widehat{NAC}$, the $BN$ and $NC$ arcs in the circumcircle of $ABC$ have the same length, since they are subtended by equal angles at $O$. It follows that the segments $BN$ and $NC$ have the same length, and since $OB=OC=R$, the line through $O$ and $N$ is the perpendicular bisector of $BC$.