Department of Mathematics Educations
EMAT 6990
by Jan White

     Given an equalateral triangle ABC and a point such that AD = BD + CD, what is the possible locus of D?

     Using GSP we can move point D around finding the different points where DB+CD = AD and it looks like some sort of curve.  Try it yourself.


 

 
    This looks like it is a circle.  Can we find the equation that satisfies the possible locus?

      Let's set up the equalateral triangle with one side along the x-axis and one vertice on the y-axis.  Now using the distance formulas we can define an equation for the locus of the points along this curve.


     Now let's set up the equations.  Using the distance formulas we get:


 

     We can use NuCalc graphing calculator to view these equations and see that the locus is indeed a curve.  Now we might see that this looks like a quadralateral inscribed in a circle and remember something about Ptolomey.


     Ptolemy's Therorem states:               AD*BC = AB*CD + BD*AC

     The product of the two diagonals is equal to the sum of the product of the two opposite sides.

     Here is a traditional proof:  On the diagonal BC locate a pont P such that the angles ADC and PDB are equal.  Since angles CAD and DBC subtend the same chord, then they are equal.  Therfore triangles ADC and BDP are similar.  So we get BD/PD = AD/AC or BD*AC=PD*AD.

       Angles CDP and ACD are equal, so triangles ADB and BCP are similar giving us the relationship CD/PC=AD/AB or CD*AB=PC*AD.  Adding the two eaquations and we that:
 
    BD*AC + CD*AB = PD*AD + PC*AD
    
    BD*AC + CD*AB = AD*BC    Q.E.D.

    Another nice use of Ptolemy's Theorem is to derive the trigonometry equations for the sums of angles.

    Let AC equal one, and define the other sides in terms of the sin and cos of the angles a and b.
                    AC = 1
                    AB = cos (a)
                    BC = sin (a)
                    AD = cos (b)
                    CD = sin(b)
                    BD = sin(a+b)

Ptolemy's Theorem:    AC*BD = AB*CD + AD*BC  and so by substitution we get:

                               1* sin(a+b) = cos (a)*sin(b) + cos (b)*sin (a)

                                    sin(a+b) = cos (a)*sin(b) + cos (b)*sin (a)