Ptolemy's TheoremDate: 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 Check out our web site! http://mathforum.org/dr.math/ |
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