Take the tour ×  Consider the following attempt at a ``thought-
free'' proof of Morley's Theorem.  Let $(x_1,y_1)$, $(x_2,y_2)$
and $(x_3,y_3)$ denote three vertices of a generic triangle.
Let $(a_1,b_1)$, $(a_2,b_2)$ and $(a_3,b_3)$ denote three more
points in the plane.  For all $\lbrace i,j,k\rbrace=\lbrace
1,2,3\rbrace$, we want to impose the condition that the line
determined by $(x_i,y_i)$ and $(a_k,b_k)$ bisects the angle from
$(a_j,b_j)$ to $(x_i,y_i)$ to $(x_j,y_j)$. Capturing this
algebraically amounts to equating two squared cosines: $$
\frac{((x_i-a_k)(x_i-a_j)+(y_i-b_k)(y_i-b_j))^2} {((x_i-
a_k)^2+(y_i-b_k)^2)((x_i-a_j)^2+(y_i-b_j)^2)}$$ $$ =
\frac{((x_i-a_k)(x_i-x_j)+(y_i-b_k)(y_i-x_j))^2} {((x_i-
a_k)^2+(y_i-b_k)^2)((x_i-x_j)^2+(y_i-x_j)^2)} $$ which
simplifies first to $$ ((x_i-a_k)(x_i-a_j)+(y_i-b_k)(y_i-
b_j))^2((x_i-x_j)^2+(y_i-x_j)^2)$$ $$ - ((x_i-a_k)(x_i-
x_j)+(y_i-b_k)(y_i-x_j))^2((x_i-a_j)^2+(y_i-b_j)^2)=0. $$ Then
expanding the left side and dividing out by the condition $$
-a_jy_i - b_jx