It is quite obvious that a quadrilateral can be inscribed in a parabola. However, can somebody provide a nice (meaning: intuitive) proof that a cyclic quadrilateral can be inscribed in it?
Further, if you can prove that it is possible to have an inscribed quadrilateral such that the diagonals are equal, then it'd be great!
On
Sure, one such example might be
\begin{align} x^2-y &= 0 \\ x^2+(y-13)^2 &= 25 \end{align}
with the intersection points being $(\pm 3, 9)$ and $(\pm 4, 16)$.
Picture courtesy of Wolfram Alpha:
I hope this helps $\ddot\smile$
Given a parabola, zoom out until it looks really thin, then draw a circle through it. Note that the two figures intersect in four points.
...Equal diagonals? Sure, just center the circle on the parabola's axis of symmetry!