Proving that the focus of a parabola lies on the circumcentre of a triangle
Short Proof:
You need two lemmas:
The foot of the perpendicular from the focus to a tangent of the parabola lies on the tangent at the vertex.That means the feet of the perpendicular to the three sides of the triangle formed by the tangents lie on a straight line, called the Simson Line which leads us to use
Simson-Wallace Theorem: The Feet of the perpendiculars from a point to the sides of a triangle are collinear iff the point lies on the circumcircle.
It now follows that focus lies on the circumcircle
Another way: Let the three lines be tangent at $t_1, t_2, t_3$ on the parabola. Then the points of intersection of these tangents, which form the vertices of the triangle are $A(at_1t_2, a(t_1+t_2))$ etc.
You can then show that these points along with $(a,0)$ form a cyclic quadrilateral, by using slopes to show that opp angles are supplementary.
)The following works, but is not very inspiring:
Take the parabola as $y=x^2$ with focus $F$ at $\bigl(0,{1\over4}\bigr)$. The tangent $t_a$ at $(a,a^2)$ has equation $y=2ax-a^2$. Intersecting $t_a$ with $t_b$ gives the vertex $C=\left({a+b\over2},ab\right)$ of the triangle. By symmetry we then have $A=\left({b+c\over2},bc\right)$ and $B=\left({c+a\over2},ca\right)$. Further computation leads to the circumcenter $$M=\left({1\over4}(a+b+c-4abc), \ {1\over8}(1+4ab+4bc+4ca)\right)\ ,$$ and the circumradius $R$ satisfies $$R^2={1\over64}(1+4a^2)(1+4b^2)(1+4c^2)\ .$$ It is now easy to verify that $F$ is lying on the circumcircle of $\triangle(ABC)$.