Ptolemy's Theorem

Date: 09/07/97 at 08:28:13
From: Anonymous
Subject: Ptolemy's Theorem

Dear Dr. Math,

I am searching for the proof for the Ptolemy's Theorem. Can you please 
give me a reference? Thanks.

Almut Breitling

Date: 09/07/97 at 18:00:53
From: Doctor Anthony
Subject: Re: Ptolemy's Theorem

Ptolemy's Theorem
------------------

If a quadrilateral is cyclic the rectangle contained by its diagonals 
is equal to the sum of the rectangles contained by opposite sides.

Draw the circle with any cyclic quadrilateral ABCD.  Draw the 
diagonals AC and BD. Suppose that <BAC is greater than <CAD. In <BAC 
draw AP, meeting BD at P, so that <BAP = <CAD.

In triangles ABP, ACD, <BAP = <CAD (construction), and <ABP = <ACD 
(angles in same segment), so the triangles are similar and:

  AB     BP 
 ---- = ----    or  AB.DC = AC.BP   ......(1)
  AC     DC 


In triangles ABC, APD   <BAC = <PAD   (i.e. <BAP +<PAC = <CAD + <PAC)

also <ACB = <ADP    (angles in same segment).  Therefore triangles are 
similar, and

   BC     AC
  ---- = ----    or   BC.AD = AC.PD   ......(2)
   PD     AD

Combining (1) and (2)

   AB.DC + BC.AD = AC.BP + AC.PD

                 = AC(BP + PD)

                 = AC.BD

and so the theorem is proved.

If ABCD is not a cyclic quadrilateral, the theorem states

  AB.CD + AD.BC > AC.BD

Construct <BAP = <CAD  and  <ABP = <ACD

In this situation <ABD does not equal <ACD and therefore <ABP does not 
equal <ABD and P does not lie on BD.

Then BP + PD > BD  and

       AB.CD + AD.BC > AC.BD

-Doctor Anthony,  The Math Forum
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