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Bretschneider's Theorem and Cyclic Quadrilaterals


Date: 11/30/2000 at 12:44:32
From: Rachel Blumberg
Subject: I need to prove Bretschneider's theorem

My Calculus III professor brought up this problem in class: Prove that 
maximizing any quadrilateral with sides ABCD means inscribing it into 
a circle.

I used Bretschneider's theorem, which states that if s = (a+b+c+d)/2 
and A and C are opposing angles in the quadrilateral, then the area of 
the quadrilateral is 

     sqrt((s-a)(s-b)(s-c)(s-d)-((cos((A+C)/2)^2)).  

When this quadrilateral was inscribed, I saw that the opposing angles 
would be supplementary, and therefore the area would be maximized 
because the cosine term would be 0.

My professor patted me on the back and said, "Now prove the theorem. 
And while you're at it, explain why defining a quadrilateral with 
supplementary opposing angles means that it is inscribable into a 
circle."

Any insight on proving this theorem would be helpful. I do not know 
where to start. Also I know that a quadrilateral inscribed in a circle 
has opposing angles that add to 180 degrees, but I don't know how to 
show the converse, that if a quadrilateral has supplementary opposing 
angles, then it can be placed inside a circle.


Date: 11/30/2000 at 16:58:43
From: Doctor Rob
Subject: Re: I need to prove Bretschneider's theorem

Thanks for writing to Ask Dr. Math, Rachel.

I once proved the cyclicity of the maximal-area quadrilateral with 
given side lengths by setting up the area as the sum of the areas of 
the two triangles formed by a diagonal, and then taking the derivative 
with respect to the length of that diagonal. When I set the derivative 
equal to zero and solved, I found that the length of the diagonal was 
such that the circumcircles of the two triangles coincided, so the 
quadrilateral was cyclic. I do not recommend doing this, however, 
since the computations were extraordinarily messy.

Using Bretschneider's Formula was a good idea. Proving that may 
turn out to be messy. I don't know what Bretschneider's original proof 
was.

To show that when opposite angles are supplementary the quadrilateral 
is cyclic is not too hard. Pick one of the two angles, say <ABC, and 
draw the diagonal AC. This forms a triangle ABC. Draw the circumcircle 
of this triangle. Then the arc AC has measure double that of of <ABC. 

Now consider line CD. This intersects the circle at point E. Draw EA. 
Then <CEA has measure half that of arc CA. Since arcs CA and AC have 
total measure 2 pi rad, or 360 degrees, <CEA and <ABC have total 
measure pi rad, or 180 degrees. 

Now since <ABC and <CDA are supplementary, and <ABC and <CEA are also 
supplementary, then <CEA = <CDA. But since the angles of triangle CDE 
add up to pi rad or 180 degrees, this implies that points E and D 
coincide, and so D lies on the circle. (This uses the fact that any 
exterior angle of a triangle is the sum of the two other interior 
angles.)

Now for Bretschneider's Formula.

Let the side lengths be a, b, c, and d. Let alpha be the measure of 
the angle between sides a and b, and beta be the measure of the angle 
between sides c and d. Now the areas of the two triangles formed by 
sides a, b, and one diagonal, and sides c, d, and the same diagonal, 
are

     K1 = (1/2)*a*b*sin(alpha)
     K2 = (1/2)*c*d*sin(beta)

Then the area of the quadrilateral is

     K = K1 + K2

As you stated, we want

     s = (a+b+c+d)/2

and then we want to show that

     K = sqrt[(s-a)*(s-b)*(s-c)*(s-d)] - a*b*c*d*cos(alpha+beta)

To eliminate K1 and K2 isn't bad, just use substitution.

     K = a*b*sin[alpha]/2 + c*d*sin[beta]/2

I'm not quite sure how to proceed from here. Good luck.

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


Date: 12/01/2000 at 13:07:06
From: Doctor Rob
Subject: Re: I need to prove Bretschneider's theorem

Hi, Rachel.

I believe I have the proof of Bretschneider's Theorem now. Cut the 
quadrilateral into two triangles using one of the diagonals. Let the 
two angles whose vertices are not on that diagonal have measures alpha 
and beta, let the lengths of the sides adjacent to angle alpha be a 
and b, let the lengths of the sides adjacent to angle beta be c and d, 
and let the length of the diagonal be p.

Then the area of the quadrilateral is the sum of the areas of the 
triangles. One triangle has base a and height b*sin(alpha), and the 
other has base c and height d*sin(beta). Thus the area of the 
quadrilateral is given by

     K = (1/2)*a*b*sin(alpha) + (1/2)*c*d*sin(beta)

Square this expression:

     K^2 = (1/4)*(a^2*b^2*sin^2[alpha]+
                  2*a*b*c*d*sin[alpha]*sin[beta]+
                  c^2*d^2*sin^2[beta])

Now use the Law of Cosines on each triangle:

     p^2 = a^2 + b^2 - 2*a*b*cos(alpha)
     p^2 = c^2 + d^2 - 2*c*d*cos(beta)

Set these equal to each other to eliminate the variable p, and isolate 
the cosines on one side of the equation:

     2*a*b*cos(alpha) - 2*c*d*cos(beta) = a^2 + b^2 - c^2 - d^2

Now square both sides of this equation, divide by 16, and bring all 
terms to the left-hand side:

     (1/4)*(a^2*b^2*cos^2[alpha]-
            2*a*b*c*d*cos[alpha]*cos[beta]+
            c^2*d^2*cos^2[beta]) - (a^2+b^2-c^2-d^2)^2/16 = 0

Add this to the equation for K^2 above, and use the fact that
sin^2(x) + cos^2(x) = 1, once with x = alpha and once with x = beta:

   K^2 = (1/4)*(a^2*b^2+c^2*d^2+
           2*a*b*c*d*[sin(alpha)*sin(beta)-cos(alpha)*cos(beta)]) -
          (a^2+b^2-c^2-d^2)^2/16

Now it is a matter of simplifying:

     K^2 = (1/4)*(a^2*b^2+c^2*d^2-2*a*b*c*d*cos[alpha+beta]) -
            (a^2+b^2-c^2-d^2)^2/16
         = (4*a^2*b^2+4*c^2+d^2-[a^2+b^2-c^2-d^2]^2)/16 -
            a*b*c*d*cos(alpha+beta)/2
         = 2*a^2*b^2+2*a^2*c^2+2*a^2*d^2+2*b^2*c^2+2*b^2*d^2+2*c^2*d^2 
            - a^4-b^4-c^4-d^4)/16 - a*b*c*d*cos(alpha+beta)/2

Now the numerator of the first fraction almost factors, and it will if 
we add and subtract 8*a*b*c*d:

     K^2 = (-a+b+c+d)*(a-b+c+d)*(a+b-c+d)*(a+b+c-d)/16 - a*b*c*d/2 -
             a*b*c*d*cos(alpha+beta)/2

Here we introduce s = (a+b+c+d)/2.

     K^2 = (s-a)*(s-b)*(s-c)*(s-d) - a*b*c*d/2 -
             a*b*c*d*cos(alpha+beta)/2
         = (s-a)*(s-b)*(s-c)*(s-d) - a*b*c*d*(1+cos[alpha+beta])/2
         = (s-a)*(s-b)*(s-c)*(s-d) - a*b*c*d*cos^2([alpha+beta]/2)

by using the cosine half-angle trigonometric identity. Finally,

     K = sqrt[(s-a)*(s-b)*(s-c)*(s-d) - a*b*c*d*cos^2([alpha+beta]/2)]

This last is Bretschneider's Formula, so we have proved what we set 
out to.

It's a bit complicated, but nothing really hard is involved (with the 
possible exception of noticing that a certain polynomial would factor 
if we added and subtracted the right thing).

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Euclidean Geometry
College Trigonometry
High School Euclidean/Plane Geometry
High School Trigonometry

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