# Talk:Steiner–Lehmus theorem

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## Removal of the "proof"

The proof below was removed for several reasons.

Generally when adding a proof make sure its notation is aligned to the notation already being used in the article, here in particular that the use of labels/variables matches the drawing. Also proofs (ideally) need to be sourced like other article content and the proof given does not match the one in the referenced literature (i.e. Coxeter). Note that privately derived unpublished "new" proofs are usually considered WP:OR, which means they are not permitted in Wikipedia (no matter whether they actually are correct or not).

Aside from these more general problems there seems to a purely mathematical issue as well, the proof makes use of a congruence theorem for triangles that does not exist (for triangles DHF, DHC). In general 2 triangles are not congruent if they have 2 sides and an angle in common. They are only congruent if the angle is located opposite of the large side (or attached to the smaller side), this condition is not met in the proof. Maybe this issue can be fixed, but in any case please do not add a proof to the article, before not only the math issues are resolved but the formal issues from above as well, i.e. the proof needs to be correct and it needs to be sourced and use a notation consistent with the rest of the article.--Kmhkmh (talk) 13:29, 8 October 2009 (UTC)

### Proof

In triangle ABC, BE & CD are equal angle bisectors, point F lies outside Triangle ABC such that BDFE is parallelogram, H is point on FC such that DH passes through E

In Triangle DHF & Triangle DHC
DH = DH(common)
since BE = CD (given) & BE = DF (opposite sides of parallelogram are equal)
therefore DF = CD
anlge DFH = angle DCH (base angles of isosceles triangle)
Therefore Triangle DHF congruent to Triangle DHC
angle DHF = angle DHC & HF = HC
In triangle EFH & triangle ECH
EH = EH (common)
angle EHF = angle EHC (proved above)
HC = HF (proved above)
In triangle EFH congruent to triangle ECH
let angle EFH = angle ECH = z
since angle DFH = angle DCH
x + z = y + z
x = y
2x = 2y
AB = AC

## A direct Euclidean proof?

In December 2010, Charles Silver of Berkeley, CA, devised a direct proof of the Steiner-Lehmus theorem, which uses only compass and straightedge and relies entirely on notions from Book I of Euclid's Elements. He submitted to The American Mathematical Monthly, but apparently it was never published. It seems to me that it should also be introduced into the section about Conway's claims of the impossibility of a proof.

It appears below, first in simplified form, then in expansion:

GIVEN: Triangle ABC with angle ABD bisecting angle ABC and angle ACE bisecting angle ACB. And the bisecting lines BD and CE are equal.

Silver's direct proof of the Steiner-Lehmus theorem

TO PROVE: Triangle ABC is isosceles.

INTUITIVE PROOF:
1. From D construct to the right a line parallel to BC which also equals BC. This right line ends at point F.
2. Similarly, from E construct to the left a line parallel to BC again, also equaling BC. This left line ends at point G.
3. There are now two parallelograms – BDFC and CEGB – found congruent by "application," or superposition.
4. Since the opposite angles of two equal parallelograms are equal, the angle at F equals angle CBD, and the angle at G equals angle BCE.
5. Therefore, twice these “half-angles” are equal, making the base angle ABC = angle ACB. Thus, triangle ABC is isosceles.

The proof relies on the assumption that Euclid defined a parallelogram as a quadrilateral figure the opposite sides of which are parallel. It obviously also relies on the controversial principle of superposition, which Euclid uses in Proposition I.4.: "“For if the triangle ABC be applied to the triangle DEF, and the point A be placed on the point D and the straight line AB on DE, then the point B will also coincide with E, because AB is equal to DE.” A Common Notion of application would then hold that "if all features of one object can be applied exactly with all features of another object, then the two objects coincide." (c.f. Oliver Byrne's The First Six Books of The Elements of Euclid, 1847, Elucidation XXV)

With those statements in mind, here is Silver's expanded proof.

GIVEN: Triangle ABC with angle ABD bisecting angle ABC and angle ACE bisecting angle ACB. And the bisecting lines BD and CE are equal.

TO PROVE: Triangle ABC is isosceles. TO PROVE: Side AB equals side AC. Thus Triangle ABC is isosceles TO PROVE: Angle ABC equals angle ACB.

PROOF:
1) Construct a circle with center D and distance BC. (By Post. 3: “To describe a circle with any centre and distance.”)
2) Through D, draw a straight line parallel to BC (that’s exterior to triangle ABC). (By Prop. 31: “Through a given point to draw a straight line parallel to a given straight line.”)
3) Point F is the intersection of the circle (by 1) and the straight line (by 2) on the other side of DE.
4) Therefore BC is parallel to DF (by 2) (and BC is equal to DF (by 1)).
5) BD is parallel (and equal) to CF (Prop. 33: “The straight lines joining equal and parallel lines (at the extremities) which are in the same directions respectively are themselves equal and parallel.”)
6) BCFD is a parallelogram. (By Def. 22., amended above)
7) CBGE is a parallelogram (By the same method as steps 1 through 6 for BCFD.)
8) Parallelogram BCFD is congruent to parallelogram CBGE. (By Def. 4.5, they coincide by 'application'. This is the order of the application. First apply the point F to G and apply the straight line FD to GE (By Prop. 31, they are both parallel and equal to BC, thus they coincide. Then the applied point D will coincide with E because both lines are equal and parallel to BC. (This is almost word for word the same as in the proof of Prop. I.4.) Next, place the point C on B and B on C (i.e., just reverse the names of the extremities of the points on the base line.) Then the applied point B will fall on the original C, and C will fall on B. Then the line from (the applied) point F to point C coincides with the original line from G to B (Prop 33 again: “The straight lines joining equal and parallel straight lines (at the extremities which are in the same directions (respectively) are themselves equal and parallel”). Likewise, apply the applied line DB on line EC. Now, all points and lines of parallelogram BCDF coincide with parallelogram BCFD. Thus, the parallelogram BCFD coincides with CBFD. Thus, the parallelogram BCDF is congruent to CBGE (by CN 4. “Things which coincide with one another are congruent (“equal”) to one another.) “And the remaining angles will also coincide with the remaining angles and will be equal to them” (quoting a line in the proof of Prop. I. 4) Hence, angle CBD equals angle BCE. But angle BCD is half of angle ACB (BD bisects the angle ACB). Similarly, angle BCE bisects angle ABC.
9) Thus, angle ABC equals angle ABC (CN 2: “If equals be added to equals, the wholes are equal”.
10) Thus, side AB equals side AC (by Prop. 6: “If in a triangle two angles be equal to one another, the sides which subtend the equal angle will also be equal.”
12) Therefore, etc. QED

Silver adds that his proof does not address the "alone" condition of Euclid's definition of the isosceles triangle from Proposition 6: “Those trilateral figures, an isosceles triangle is that which has two of its sides alone equal.” He concludes: "Using this definition, special precautions could be taken to ensure that the legs of the triangle each differ in length from the base, but it seems preferable simply to accept the standard notion that an isosceles triangle does not exclude it from being equilateral, rather than asserting that an isosceles triangle must have only two sides equal."

Silver refers to the proof as the Transamerica proof of Steiner-Lehmus, owing to the resemblance of the figure above with its parallelogram "wings" and the iconic Transamerica Pyramid in San Francisco, CA. — Preceding unsigned comment added by Skjames44 (talkcontribs) 17:00, 21 June 2011 (UTC)

Seems to me this is one of those proofs where the diagram "lies" by assuming that which is to be proved. While we know that DF and EG are parallel, we don't know that they are collinear. Hence we may need to distort BCDF (while keeping all side lengths constant) in order to "apply" it to CBGE. This will make the applied BCDF congruent to CBGE, which is what the above proof seems to be using, but the original BCDF need not be.
Meanwhile I don't understand what disqualifies a proof as "direct." If a proof shows directly that xy and yx, is the inference that x = y allowed as part of a direct proof? --Vaughan Pratt (talk) 17:26, 21 August 2011 (UTC)

## Clean up

I clean out the proofs for several reasons:

• Proofs are usually not appropriate in math articles on theorems (see Wikipedia:Manual_of_Style_(mathematics)#Proofs)
• WP is not a venue or a means of promotion for unpublished material
• If you nevertheless want to offer readers access to a proof you can do that by providing an external link containing a proof. Alternatively in particular if you want to offer a proof under CC licence you could write them up on wikibooks or wikiversity site and link it here. Wikibooks already has a specific project 'book of proofs' exactly designed for that purpose.

--Kmhkmh (talk) 03:46, 9 August 2011 (UTC)

## Restoration of a valid addition

The reason I restored the reference to a 'direct proof' is that it exists in a respected publication and is left to the reader to decide if it qualifies as a 'direct proof.' Since this is a published work, please leave any reference to it entact. Thank you. — Preceding unsigned comment added by MathMan1010 (talkcontribs) 06:35, 27 August 2011 (UTC)

The proof at the added reference does not seem substantially different from the one given at [1] under "algebraic proof" (and the "Sayre-Reitsma Direct Proof" is not in what most persons would characterize as a "respected publication"). Furthermore, it does not address Conway's objections as stated at: [2]. — Myasuda (talk) 13:34, 21 September 2011 (UTC)