0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ There is a tag for proof verification and you might want to add it. $\endgroup$ – cr001 Nov 18 '15 at 4:24
  • 1
    $\begingroup$ 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. $\endgroup$ – cr001 Nov 18 '15 at 4:28
  • $\begingroup$ @cr001 I made the edit that you suggested - replaced "\phi" with "\theta". Thanks. $\endgroup$ – user74973 Nov 18 '15 at 4:56
  • $\begingroup$ @cr001 I added the tag that you suggested. I did not know about this option. $\endgroup$ – user74973 Nov 18 '15 at 4:58
1
$\begingroup$

I have a contradiction which is slightly simpler.

enter image description here

Suppose that the said median and the angle bisector are not distinct. Then, the right angled triangle will have lengths as shown.

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user74973 Nov 18 '15 at 12:13
  • $\begingroup$ I should complement you on your demonstration. Very Nice! $\endgroup$ – user74973 Nov 18 '15 at 12:13
  • $\begingroup$ @user74973 It's sad & also funny that the angle bisector theorem is being excluded in many schools/ institutes. First of all, the proof of it is not that difficult to derive. Secondly, it is very powerful in certain occasions. Without it, one might have to go thro' many extra steps before the desired can be attained. Regarding your book, I suggest introducing the theorem 'indirectly' as:- 1) After similar triangles being taught, 2) introduce the angle bisector environment with the needed construction included. 3) Ask the readers to prove the needed ratio. 4) Solve problems requiring the ratio. $\endgroup$ – Mick Nov 18 '15 at 15:16
  • $\begingroup$ The books to which I refer to it as "The Triangle-Bisector Theorem." I think it is ridiculous that it is not taught. The demonstration of it is a clever use of auxiliary lines to verify geometric propositions. Except for using it to verify other important propositions, I haven't seen any nice examples using it. (By the way, I would describe what I am doing as writing several chapters on various topics in geometry. It is to be used for a supplement for a college course in geometry - a course in which students learn to prove statements.) $\endgroup$ – user74973 Nov 18 '15 at 15:30
  • $\begingroup$ Having said that, do you agree that the argument I provided is lucid? $\endgroup$ – user74973 Nov 18 '15 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.