Talk:Angle bisector theorem

Is this: BD:DC = AC:AB true, or is it really this: BD:DC = AB:AC? Loggie 22:20, September 3, 2005 (UTC)

You're right, it should be BD:DC = AB:AC. Well spotted. -- Jitse Niesen (talk) 23:44, 3 September 2005 (UTC)

Discussion about readding Euclid's original proof in this article - see Euclid's proof section below

I added the limited but highly accessible proof in a hurry and being limited to mobile I was unable to cite properly. I don't know how to cite the book I have with me, but it's available on proofwiki. I think it should be included because it adds quality and accessible simplicity to the article. Showar (talk) 22:43, 23 April 2020 (UTC)

???????

If only point D is moved along BC, ratio AB : AC is unchanged. So, I think it is all nonsense and should be deleted. --84.118.81.7 (talk) 09:22, 18 May 2014 (UTC)

D here is specifically the point of intersection between BC and the line internally bisecting the angle A of the the triangle, if you are actually confused.GenericName784 (talk) 11:39, 21 April 2020 (UTC)

Euclid's proof

Is it worth mentioning that this was proved in Euclid's The Elements Book VI Proposition 3? Admittedly it doesn't solve the general case, but it's notable all the same. If we don't want to actually include it completely, we can just link to the version on ProofWiki. --Matt Westwood 22:17, 13 August 2011 (UTC)

So did Euclid fundamentally create this theorem? 131.107.65.118 (talk) 01:14, 11 October 2011 (UTC)

Should that make a difference? --Matt Westwood 05:33, 11 October 2011 (UTC)

Sorry, I didn't meant to seem like I disagreed with including it. I think it is notable and should be included, I was just wondering where the orgins of the Angle Bisector Theorem lie, did they start in that book? — Preceding unsigned comment added by 50.135.192.196 (talk) 03:54, 12 October 2011 (UTC)

According to Thomas L. Heath (I just looked it up), something similar can be found in Aristotle, somewhat earlier than Euclid. So: no. Although Euclid may have polished it and extended it somewhat. --Matt Westwood 18:23, 12 October 2011 (UTC)

Okay thanks! :) 50.135.192.196 (talk)

additional sources

• https://www.jstor.org/stable/3610280
• https://www.jstor.org/stable/27957506
• https://books.google.de/books?id=9grsxFZUci8C&pg=PA4
• http://jwilson.coe.uga.edu/MATH7200/Sect4.2.html

The first triangle should not look like a right triangle

It confuses the less competent reader. If an angle is not a right angle, make it obviously not so. Tuntable (talk) 05:28, 15 December 2019 (UTC)

I don't see anything looking like right triangle. Which graphic are you referring to?--Kmhkmh (talk) 10:58, 15 December 2019 (UTC)

Deleted Proof

I added this proof but was deleted:

Let $h$ be the height of the triangles on base BC.

${\frac {[ABD]}{[ACD]}}={\frac {\frac {BD.h}{2}}{\frac {CD.h}{2}}}={\frac {\left({\frac {AB.AD.\sin \angle BAD}{2}}\right)}{\left({\frac {AC.AD.\sin \angle CAD}{2}}\right)}}$

$\because \sin \angle BAD=\sin \angle CAD$

$\implies {\frac {BD}{CD}}={\frac {AB}{AC}}$

Shubhrajit Sadhukhan (talk) 05:18, 25 May 2020 (UTC)

Triangle not isosceles

It would be good if the intro triangle was obviously skewed, and not looking isosceles. That would make it clear what the ratios are. isosceles is a trivial case. Not important to a sophisticated mathematician, but very important to a less worthy student.Tuntable (talk) 00:51, 14 October 2020 (UTC)

proof using similar triangles

I'd suggest to replace the animation by a fixed image. Animations are usually not well suited for proofs which people read at different speeds and may want to mull over a specific argument for a second. In such cases it is extremely annoying if the illustration constantly changes.--Kmhkmh (talk) 01:12, 26 February 2024 (UTC)