This is a follow-up question on Geometry question with perpendicular lines and angle bisectors in triangle.
The construction is the same as that question:
For any triangle ABC
1) Draw angle bisector of angle BAC. Draw perpendicular bisector of BC. Mark the intersection point as D.
2) Draw altitudes from D to AB and AC. (Extend line as necessary).
Last time I asked BE = CF and it was solved in that post. However, I am trying to prove another observation I found while messing around in Geogebra:
- One of point E, F is always outside the triangle ABC.
I still have no idea on how to prove it. Using my best coordinate geometry I am not able to get much progress, as it is not the best way to deal with distances and angle bisectors, obviously. (lol) Hope I can get some hint.
Thank you,
Gareth
Best Answer
It's true unless $E$ and $F$ coincide with $B$ and $C$, respectively. But we can call this a degenerate case (it occurs when $AB=AC$). Note that $D$ lies on the circumscribed circle of $\triangle ABC$. This is a well-known fact for your construction. Thus, $$\angle DBA+\angle DCA=180^{\circ}$$ which means that one of those angles has to be $\geq 90^{\circ}$, say $\angle DCA$. This will force the foot of the perpendicular from $D$ to $AC$ to lie on its extension, i.e., outside the triangle. And it means that the other foot lies internally on the side.
Update: Responding to OP's comment, to show that $D$ lies on the circumscribed circle, let the perpendicular bisector meet the circle at point $D_1$. Our aim is to show $D_1\equiv D$. We have $\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}} BD_1=CD_1 \Rightarrow$ arcs $\arc{BD_1}=\arc{CD_1}\Rightarrow \angle CAD_1 =\angle BAD_1$ so $D_1$ is also on the angle bisector, thus $D\equiv D_1$.