# The median and angle bisector from a vertex being distinct

I have an argument here for the proposition that the median and the angle bisector from an acute angle in a right triangle are distinct line segments. I would appreciate comments on it. (I can include a diagram that may be compiled on LaTeX.)

Proposition

$\triangle{ABC}$ is a right triangle with its right angle at $A$. The median and the angle bisector from either $B$ or $C$ are distinct line segments.


Demonstration

$P$ is any point on the leg $\overline{AB}$, and $\theta = {\mathrm{m}}\angle{CPB}$. According to the Pythagorean Theorem, $\bigl\vert \overline{BC} \bigr\vert > \bigl\vert \overline{CP} \bigr\vert$. Since $\sin\theta \leq 1$, \begin{equation*} \frac{\sin\theta}{\overline{BC}} < \frac{1}{\overline{CP}} . \end{equation*} If $\overline{CP}$ were both the median and the angle bisector of the given triangle from $C$, and if ${\mathrm{m}}\angle{ACP} = \phi = {\mathrm{m}}\angle{PCB}$, according to the Law of Sines, \begin{equation*} \frac{1}{\overline{CP}} = \frac{\sin\phi}{\overline{AP}} = \frac{\sin\phi}{\overline{BP}} = \frac{\sin\theta}{\overline{BC}} . \end{equation*} This is a contradiction.

• There is a tag for proof verification and you might want to add it. – cr001 Nov 18 '15 at 4:24
• In your last part of the first line and in your last equality of the second last line, the $\sin(\phi)$ probably should be $\sin(\theta)$ and other than that I think the proof is good. – cr001 Nov 18 '15 at 4:28
• @cr001 I made the edit that you suggested - replaced "\phi" with "\theta". Thanks. – user74973 Nov 18 '15 at 4:56
• @cr001 I added the tag that you suggested. I did not know about this option. – user74973 Nov 18 '15 at 4:58

By angle bisector theorem, $\dfrac {a}{x} = \dfrac {b}{x}$. Consequently, a = b. This is a contradiction because the hypotenuse should be the longest in a right angled triangle.
• I am writing a chapter for a course in Geometry, and citing the Angle-Bisector Theorem would be out of order. I do have the Law of Sines (and the Law of Cosines) preceding this proposition. I am looking for an elementary demonstration for this proposition. Do you agree that the demonstration that I provided is correct? By the way, I do have code in TikZ that I could include. Do you think that I should include it? It would make my post look really messy. – user74973 Nov 18 '15 at 12:13