Geometry Stumper

[Albert Chan]

There is an even simpler Mathematica prove, without finding out slopes

Code:

In[1]:= pointE = {x, y} /. Solve[y == ae*x == be*(x - 1), {x, y}] // First;
In[2]:= pointD = pointE /. {ae -> -1/ae, be -> -1/be} // Simplify;
In[3]:= pointC = {k, 1};
In[4]:= Det[{pointD - pointC, pointE - pointC}] // Together
           2          2        2                   2             2
        -ae  - be - ae  be + be  - ae k + be k + ae  be k - ae be  k
Out[4]= ------------------------------------------------------------
                                          2
                                 (ae - be)
In[5] = % /. {ae^2 -> 1-2*k*ae, be^2 -> 1-2*(k-1)*be} // Expand           
Out[5]:= 0

Above only used the half-angle formula: tan(2x) = 2 tan(x) / (1 - tan(x)^2)

tan(Angle CAB) = 1/(k-0) = 2*ae / (1 - ae^2), thus ae^2 = 1-2*k*ae
tan(Angle CBA) = 1/(k-1) = 2*be / (1 - be^2), thus be^2 = 1-2*(k-1)*be

since determinant = 0, slope DC = slope CE, thus DCE is straight line