algebra-precalculusanalytic geometrygeometry
Show that BE is the perpendicular bisector to AC.
I tried to prove this through Pythagoras, but the answer I got did not prove it was at a right angle, and therefore said it was not the perpendicular bisector. I'm probably making a silly mistake, so any advice would be helpful!
It is easier to prove the following equivalent version:
Let $N$ be the point of intersection between the circumcircle of $ABC$ (centered at $O$) and the angle bisector from $A$. Then $N$ lies on the perpendicular bisector of $BC$.
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 $BC$.
Using the theorem about the internal bisector, we get $$\frac{BD}{DC}=\frac{c}{b}$$ so we get $$BD=DC\cdot \frac{c}{b}$$ Using that $$DC=a-BD$$ We get $$BD=(a-BD)\cdot \frac{c}{b}$$ and we obtain $$BD(1+\frac{c}{b})=\frac{ac}{b}$$ Can you finish?
Best Answer
A meta-analysis:
The calculation with Pythagoras' theorem should in fact work for the points indicated.
If done correctly it has to be consistent with any correct method.
I hope the result of the Q & A is not to treat the posted solutions as correct (maybe because there are several of them that all support the problem statement, sound authoritative, are posted by high-ranked math.SE users, or you checked the reasoning in detail, or whatever) but simultaneously disregard the original computational evidence that another equally valid principle seems to give a different answer. Tracing back to locate the source of the inconsistency should be worthwhile.