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Steiner-Lehmus Theorem


Date: 03/24/2000 at 09:16:40
From: Karri Chandra Sekhar
Subject: Geometry

1. If in a triangle the lengths of two angle bisectors are equal, how 
can we prove that it is an isosceles triangle?

(For this one, I tried and got an indirect solution. That is, I 
started at an assumption that resulted in a contradiction, so I have 
to change my assumption and try again to arrive at the correct 
solution. With this I am not satisfied. I need a direct solution. Can 
someone help me, please?)

2. There is a theorem that says that in a cyclic quadrilateral, the 
sum of opposite angles is 180 degrees. Is the converse of this theorem 
also true? That is, if in a quadrilateral, opposite angles add up to 
180 degrees, can we prove that it is a cyclic quadrilateral?

(For this one, it is enough to prove the converse of the following 
theorem: in a circle, if an arc forms 2x degrees at the center, it 
subtends an angle x on the circumference (i.e. at any point) above 
that arc.

Kindly help me in solving these two problems.

Thanking you,
Chandra Sekhar Karri


Date: 03/24/2000 at 16:24:00
From: Doctor Rob
Subject: Re: Geometry

Thanks for writing to Ask Dr. Math, Karri.

You can prove the theorem directly by using the formula for the length 
of the angle bisectors. Let the lengths of the sides of the triangle 
be a, b, and c. Then let the bisectors of the angles opposite the 
sides of lengths a and b have lengths t_a and t_b. Then the formulas 
for t_a and t_b from the Dr. Math FAQ page of Triangle Formulas 
(scroll down to the scalene triangle) are: 

     t_a = sqrt(b*c*[1 - a^2/(b+c)^2])
     t_b = sqrt(c*a*[1 - b^2/(c+a)^2])

  http://mathforum.org/dr.math/faq/formulas/faq.triangle.html   

Set these equal to each other. Square both sides, and multiply by 
(b+c)^2*(c+a)^2 to clear all fractions. (This factor can't be zero, so 
this is allowable.) Bring everything over onto one side of the 
equation, combine like terms, and factor the resulting polynomial. 
When I did this, I got

     c*(b-a)*(a+b+c)*(a^2*b+a*b^2+3*a*b*c+a*c^2+b*c^2+c^3) = 0

Now since all of a, b, and c are positive, the only way this equation 
could be true is if b - a = 0, that is, a = b.

The converse of the theorem above, "In a circle if an arc forms 2x 
degrees at the center, it subtends an angle x on the circumference 
(i.e. at any point) above that arc," would say, "If the vertex of an 
angle of measure x lies on a circle, then the arc its subtends has 
measure 2*x at the center."

Draw the angle and the circle. From the vertex of the angle, draw a 
diameter of the circle, and from the points of intersection of the 
sides of the angle and the circle, draw radii of the circle. Now the 
angle is the sum (or difference) of two angles a and b made by the 
sides of the angle and that diameter. Thus x = a + b, or x = a - b, or 
x = b - a, depending on the relative position of the diameter and the 
two sides of the angle. Each of the two triangles so constructed is 
isosceles, because two of their sides are radii of the circle, hence 
equal. Thus the base angles are equal, and so equal to a or b, 
respectively. Then the external angles at the center formed by the 
diameter and each of the radii are 2*a and 2*b, because they are the 
sum of the interior angles at the other two vertices of each triangle, 
which are both equal to a or b. Then the central angle subtending the 
arc is 2*(a+b) or 2*(a-b), or 2*(b-a), and so equal to 2*x.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   


Date: 04/22/2000 at 05:38:50
From: chandra karri
Subject: Re: Geometry

Hello Dr. Math,

First of all, let me thank you for your kind reply.

As to the proofs of the problems I posed to you, in the case of first 
one, I need a proof based on pure geometry only (there should be no 
involvement of trigonometry.) You have supplied a proof that used 
expressions for the lengths of angle-bisectors. If possible, please 
send a direct proof based on existing theorems of geometry.

The proof of second problem ... I could not follow it or visualize it. 
So if possible, please send the same proof with some illustrations.

Then finally, I have a doubt about the definition of a point in 
geometry. How can we define a point in geometry?

That's all for now,

Thankingly,
CS Karri.


Date: 04/22/2000 at 11:53:42
From: Doctor Rob
Subject: Re: Geometry

The first problem is actually a famous theorem. It has a name: the 
Steiner-Lehmus Theorem. All the usual proofs are by contradiction, 
like your original one. The proof I gave using the formula for the 
lengths of the angle bisectors is direct. It is not trigonometric. It 
does use algebra, but not trigonometry. I'm sorry if it does not 
satisfy your stringent restrictions on the method of proof you will 
allow. 

The formula for the angle bisectors is a direct consequence of 
Stewart's Theorem, which says: Let d be the length of a line segment 
through C that divides the opposite side of length c into segments of 
length m and n adjacent to the sides of length a and b respectively. 
Then n*a^2 + m*b^2 = c*(d^2 + m*n). (See crudely drawn picture below.)

              C
              o
             / \\
            /   \ \
           /     \  \
        a /     d \   \ b
         /         \    \
        /     m     \  n  \
       o-------------o------o
     B          c             A

Of course m + n = c. When the line segment of length d is the bisector 
of angle C, then n/b = m/a, which is a well-known theorem of geometry. 
Putting these together, you get

     c = m + n,
       = m*(1 + b/a)

     m = a*c/(a+b)

     n = b*c/(a+b)

     a^2*b*c/(a+b) + a*b^2*c/(a+b) = c*d^2 + a*b*c^3/(a+b)^2

     a*b*c*(a+b)/(a+b) = c*(d^2+a*b*c^2/(a+b)^2)

     a*b = d^2 + a*b*c^2/(a+b)^2

     a*b*(1-c^2/(a+b)^2) = d^2

That gives you the formula for the length of the angle bisector 
t_c = d of angle C as

     t_c = sqrt[a*b*(1-c^2/(a+b)^2)]

Relabeling the sides gives the length of the angle bisectors of angles 
A and B as

     t_a = sqrt[b*c*(1-a^2/(b+c)^2)]
     t_b = sqrt[c*a*(1-b^2/(c+a)^2)]

Thus, using just a little algebra, the Steiner-Lehmus Theorem is a 
consequence of Stewart's Theorem and other known theorems of plane 
geometry.

There is also a fair amount of literature about this problem. Here are 
a few references turned up by a search:

Court, College Geometry (1952), Barnes and Noble, p.72.

H. S. M. Coxeter, Introduction to Geometry Second Edition, p. 420: He 
mentions that J. A. McBride has asserted that more than 60 proofs have 
been given.

Coxeter and Geitzer, Geometry Revisited (1967), Random House, p. 15: 
Lehmus's proof of 1848.

J. Fellman, Mathematics Student Journal (1962), p. 5.

M. Gardner, Scientific American 204 (1961), pp. 166-168: with a follow 
up about 2 months later; the article in the book follows even this up 
in a footnote. (Book unknown).  [I believe the book is titled New 
Mathematical Diversions: More Puzzles, Problems, Games, and Other 
Mathematical Diversions  -Dr. TWE]

G. Gilbert and D. MacDonnell, American Mathematics Monthly 70 (1963), 
pp. 79-80.

Gilbert and MacDonnell, in the M.A.A. "Selected Papers on Geometry" 
(1979), pp. 127-8; reprinted from the American Mathematical Monthly 70 
(1963), pp. 79-80.

A. Henderson, Scripta Mathematica (1955), pp. 223-232, 309-312. (I 
have not seen this article, but the title is intriguing: The 
Lehmus-Steiner-Terquem problem in global survey.)

M. Levin, Mathematics Magazine 47 (1974), pp. 87-89: he mentions that 
Rougevain was the first to publish a proof (1842).

J. Malesevic, Mathematics Magazine (1970), pp. 101-102.

J.A. McBride, Edinburgh Mathematical Notes 33 (1943), pp. 1-13.

H.C. Overley, School Science & Mathematics (unknown date).

A. Posamentier, Advanced Euclidean Geometry (1979), p. 72-75: 4 
proofs.

K. Seydel and C. Newman: Unknown, pp. 72-75 (after 1982; looks like an 
MAA journal, next article was a tribute to Beckenbach).


For the second problem, here is a rough diagram of the case where the 
diameter lies between the sides of the angle:

             ,,--'--..
         _,-'        _,-o B
        /        _,-' /   \
       .     _,-'   /      .
       . _,-'     /        .
     A o---------o---------o D
       .`.      O|         .
       .  `.     |         .
        \   `.   |        /
         `-._ `. |    _,-'
             ``--o--''
                  C

Here points A, B, C, and D lie on the circle. <CAB is the angle x you 
started with. O is the center of the circle. OA = OB = OC because they 
are radii of the circle. That means that <OAB = <ABO = (1/2)*<DOB, and 
<CAO = <OCA = (1/2)*<COD (since any external angle of a triangle is 
equal to the sum of the other two internal angles). Thus

     x = <CAB
       = <CAO + <OAB
       = (1/2)*(<COD+<DOB)
       = (1/2)*<COB

     2*x = <COB

The cases where the diameter AD lies outside the angle <CAB are much 
the same, but you have to subtract the angles <CAO and <OAB to get 
<CAB, and likewise you have to subtract the angles <COD and <DOB to 
get <COB.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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