Geometry Stumper

[Albert Chan]

Another proof that CDE lies on a straight line, Mathematica, without using Solve:

From analytic geometry thread, learn about properties of excircles.

Rotate the triangle, so BC lies on line x=1, AC on t2 radian line, AB on t3 radian line.
For ABC with inscribed unit circle, these are the coordinates: (d is calculated using symmetry, then rotation)

PHP Code:

c = {1, Tan[t2/2]}; 
e = {1 - Tan[t2/2] Tan[t3/2], Tan[t2/2] + Tan[t3/2]}; (* excircle center *)
d = {{Cos[t2], -Sin[t2]}, {Sin[t2], Cos[t2]}} . (e /. {t2 → t3-t2, t3 → 0-t2}); 


Prove is now trivial:

PHP Code:

Simplify[Det[{d-c, e-c}]] ==> 0 


It can also confirm by distance, with slight help from our end (avoid square root):
If CDE on straight line, distance wise, EC + CD = ED
Square both side: EC² + CD² + 2 √(EC² CD²) = ED²
Rearrange, square again: 4 EC² CD² = (ED² - EC² - CD²)²

PHP Code:

dSqr[pt_] := Simplify[Dot[pt, pt]]
{ec2, cd2, ed2} = Map[dSqr, {e-c, c-d, e-d}];
Simplify[4 ec2 cd2 == (ed2 - ec2 - cd2)^2] ==> True