Steiner-Lehmus Theorem:
Prove if two angle bisectors of a triangle are equal, then the triangle
is isosceles (more specifically, the sides opposite to the two angles
being bisected are equal).
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Coxeter attributes this proof to "a letter from H.G. Forder"
(Coxeter's Introduction to Geometry)
ABC is the triangle, BY, CZ the bisectors of B, C, with Y on AC, Z on AB.
Assume angle B < angle C.
Let U on AB be such that angle UCZ = half angle B [3]. angle UCZ < angle
ACZ, so U is between A and Z.
Let V on BU be such that BV = CU [1]. Angle UBC < angle UCB, so UB > UC. V
is therefore between U and B.
Let W on BY be such that VW is parallel to UC [2]. Then W and Y are on
opposite sides of the line UC.
In triangles CUZ
BVW,
BV = CU (by [1]), BVW = CUZ (by [2]), VBW = UCZ (by [3]). So they are
congruent. Hence CZ = BW < BY.
We have proved angle B < angle C => CZ < BY
i.e. CZ >= CY => angle B >= angle C
Similarly CY >= CZ => angle C >= angle B
Therefore CY = CZ => angle B = angle C, so ABC is isosceles.
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120. A GEOMETRICAL DOODLEBUG [W. T. Williams & G. H. Savage, 1946]
CAN you score a direct hit on this innocent-looking, but rather deadly,
bit of Geometry---in other words, give a strictly Euclidean proof
(without recourse to {\it reductio ad absurdum} methods) of the following :--
``If the angle-bisectors, BE and CF, of a triangle ABC are
equal, the triangle is isosceles.''
Some years ago, a widely circulated weekly invited a direct
proof from its readers; but though several indirect proofs were
sent in, a direct proof was not forthcoming. Nor, to the best
of our knowledge, has any such proof appeared hitherto.
However, one will be found on page 129. {to which we now turn}
120. A GEOMETRICAL DOODLEBUG
A
/ /|\
/ / | \
/ / | \
/ / | \
/ / | \
/ / | \
/ / | \
/ / | \
/ F / | \ E
G ----------------------------/---------|---------\
\ / \ | / \
| \ / \ | / \
\ \ / \ | / \
| \ / \ | / \
| \ / \|/ \
\ \ / / \ \
| \ / / I \ \
| \ / / \ \
\ \ / / \ \
| / / \ \
| / \ / \ \
\ / \ / \ \
| / / \ \
| / / H \ \
\ / / \ \
| / / \ \
| / / \ \
\ / / \ \
| / / \ \
\// \\
B ------------------------------------------------------------- C
On BE describe a triangle GBE, congruent with AFC. Let the bisector
of the angle G meet BE in H. Join I, the intersection of BE and
CF, to A. Then IA bisects angle A. Join AG.
Since angle BGE = angle A, B, G, A, E are concyclic.
Therefore angle ABE = angle AGE
Therefore angle AIE (= angle ABI + angle BAI) = angle AGE + A/2 =
= angle AGE + (angle BGE)/2 = angle AGH
Therefore A, G, H, I are concyclic.
But GH = AI (bisectors of corresponding angles in two congruent
triangles). Therefore AG is parallel to HI.
Then C/2 = angle BEG = angle AGE (alternate) = angle ABE = B/2.
Therefore C = B, therefore ABC is isosceles.
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An algebraic proof: [Baptist], [Specht, D.8, D.68, D.69, D.70]
Thm of Stewart: Length of a traversal CP
C
/ / \
b / /t \ a
/ / \
A----P----------B
m n
t^2 = m a^2 / c + n b^2 /c - m n = (m a^2 + n b^2)/(m + n) - m n
For the angle bisector at point C we know the ratio m : n = b : a.
With c=m+n we can deduce m = bc/(a+b) and n = ac/(a+b) and
t_c^2 = ab (1 - c^2/(a+b)^2) = ab(a+b+c)(a+b-c)/(a+b)^2
Now we have equal angle bisectors t_a = t_b that is
0 = t_a^2 - t_b^2
= c(a+b+c)( b(b+c-a)/(b+c)^2 - a(a+c-b)/(a+c)^2 )
= c(a+b+c)(b-a)( (a+b+c)(ab+c^2) + 2abc )/( (b+c)^2 (a+c)^2 )
Only the third factor be zero for positive a, b, c.
Therefore we have a = b.
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A proof by contradiction using trigonometry: M. Simon (1906)
Let ABC be a triangle and let AD = BE be the (internal) angle
bisectors of A and B. For the sides I use the names a:BC, b:AC and
c:AB. For the area of ABC I use O.
The area of triangle ADB is 0.5*c*AD*sin(0.5*A).
The area of triangle ADC is 0.5*b*AD*sin(0.5*A).
Triangles ADB and ADC add up to ABC, so AD*(b+c)*sin(0.5*A) = 2O. [1]
In the same way you can find BE*(a+c)*sin(0.5*B) = 2O. [2]
From [1] = [2] and AD = BE we derive:
(b+c)*sin(0.5*A) = (a+c)*sin(0.5*B)
b*sin(0.5*A) - a*sin(0.5*B) = c(sin(0.5*B)-sin(0.5*A)) [3]
Now suppose that angle B > angle A.
Then sin(0.5*B)>sin(0.5*A) (both half angles are < 90 deg), so the
righthand side of [3] is positive.
However cos(0.5*B) angle A. Likewise, we can't have angle
A > angle B.
So angle A = angle B and triangle ABC is isosceles, so the
Steiner-Lehmus theorem is proven.
- Doctor Floor, The Math Forum
http://forum.swarthmore.edu/dr.math/
http://forum.swarthmore.edu/dr.math/problems/mcintosh1.28.99.html
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A proof using SSA: F. G. Hesse (1842)
C
/ \
/ \
F___________E_____D
\ / \ / \
\ / G \
\ / / \ \
\ / / \ \
\ / / \ \
A/_______________\B
We consider given that ABC is a triangle,
AD and BE bisect angles A and B such that AD=BE.
Construct F such that AF=AE and DF=AB.
Let G be the intersection of AD and BE.
We find ADF and EBA are congruent, so pi/2
From this we can conclude the congruence of BAF and FDB despite only
having SSA, because the unknown angles must be acute. So DB = AF = AE
and thus 1/2 Coxeter's Introduction to Geometry discusses this theorem,
> refers to it as the Stiener-Lehmus theorem, and says there are
> over 60 proofs. He gives a simple, elegant proof of the more general
> theorem that the larger of two angle bisectors intersects the
> smaller side. The proof is independant of the parallel
> postulate, and thus is also valid in hyperbolic geometry.
I've often responded to questions about why this theorem is so hard to
prove by saying that it isn't true if you allow the various lengths
involved to be in more general fields, like the complex numbers. After
someone asked for details about this the other day, I took the trouble
to work everything out, and found to my surprise that it isn't even
true when you allow them to be negative reals, which makes the
situation much easier to understand.
Here's what's going on. The lengths of the A and C angle
bisectors turn out to be
root{bc(a+b+c)(b+c-a)} root{ab(a+b+c)(a+b-c)}
---------------------- and ---------------------- ,
b+c a+b
so the equality we're given is
bc(b+c-a).(a+b)^2 = ab(a+b-c).(b+c)^2
or
(c-a){ ac^2 + (a^2+3ab+b^2)c^2 + b^2(a+b) } = 0.
Now algebraically, this does not imply that c=a, because
it might be the second factor that vanishes.
However, if a,b,c are positive, then so is this second factor,
and so with this assumption we CAN deduce c = a, but only by a proof
that somehow distinguishes positive numbers from negative ones. Now
the most common type of proof by a chain of equalities, such as:
"This length equals that length, therefore those two triangles
are congruent, so this angle equals that angle, whence ... whence
ultimately AB = BC."
cannot possibly do this, because it would work for all values of the
variables, and it's just not possible to deduce that c=a from
(c-a)(ac^2 + (a^2+3ab+b^2)c + b^2(a+b)} = 0.
The 60 proofs you mention avoid this by using order properties
of the reals, typically:
"Suppose AB > BC. Then this angle exceeds that angle, whence ...
whence ultimately the first angle-bisector is greater than the
second. Now suppose instead that AB < BC : this implies similarly
that the second is greater than the first. But we are given that
they are equal, so the only possibility is that AB = BC."
We can actually smoothly vary the parameters of a triangle in such
a way that one of its "edgelengths" changes sign and get a
"counterexample" to the Steiner-Lehmus theorem. Here's how:
C
/ \
E D
/ I \
/ \
A--> F <--B
Fix C, but let A and B move freely on the x-axis. Then if
they move so as to pass right through each other:
C
/ \
/ \
/ \
/ \
<--B F A-->
I
D E
it's natural to say that c has changed sign. As they moved, the
incenter I moved smoothly down, and crossed the x-axis as they
passed through each other, so that it became an excenter of the
final inside-out triangle.
The angle-bisector segments AD,BE,CF of the original triangle
also vary smoothly during this process, and I've drawn the final
positions that they get into in the above picture. Now take a = b = 1;
then the above equation for c becomes
c^2 + 5c + 2 = 0,
which is satisfied by c = (-5 + root17)/2, which is about -.44 ;
then with this value of c ALL THREE of these segments are equal,
but the triangle is only isosceles, not equilateral!
If there were a proof of the Steiner-Lehmus theorem of the above
equality-chasing type, it would continue to work when we varied the
triangle smoothly in this way, and so would prove this triangle
equilateral, which it isn't. So there's no such proof!
John Conway
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==> geometry/bisector.s <== [rec.puzzles Archive (geometry)]
Unfortunatly this proof is not valid. It covers only one case.
Can you find the flaw?
PROVE: XE.
Then ang(PDX) < ang(QEX)
Now considering triangles BXD and CXE,
the last condition requires that
ang(DBX) > ang(ECX)
OR ang(XBC) > ang(XCB)
OR XC > XB
Thus our assumption leads to :
XC + XD > XE + XB
OR CD > BE
which is a contradiction.
Similarly, one can show that XD < XE leads to a contradiction too.
Hence XD = XE => CX = BX
Therefore triangle BXC is an isosceles triangle.
We finish with geometry/bisector.s <==
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"Dozenten der Mathematik an der Berliner Universita"t"
Lehmus, Daniel Christian Ludolf.
Privatdozent. 18.12.1813 f"ur Mathematik, ausgesch. Ostern 1815 in den Schuldienst
*3.7.1780 Soest, +18.1.1863 Berlin. -
L.: ADB 18,147.
S.: Poggendorff 1.
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--
mailto:Torsten.Sillke@uni-bielefeld.de
http://www.mathematik.uni-bielefeld.de/~sillke/