Geometric proof that the product of the $x$-intercepts equals the $y$-intercept for a monic quadratic I know you can prove that the product of the roots of the monic quadratic $x^2+a_1x+a_0$ equals the $y$-intercept $a_0$ by comparing its coefficients to the coefficients of $(x-m)(x-c)$ where $m$ and $c$ are the roots. So $a_0 = mc$. This is how Vieta's formulas are derived.
However, I was wondering if there was a geometric proof of why this is true.
I drew the diagram below:

In the diagram the roots are $(m, 0)$ and $(c, 0)$ while the y-intercept is $(0, b)$. I also drew the point directly above the vertex of the parabola (the midpoint of the roots) and created a few triangles. I tried using Stewart's Theorem on some of the triangles but couldn't seem to get the desired result that $b = mc$.
Can anyone provide some insight on how to prove this fact geometrically? Would I need to also draw the focus and directrix and do some geometry using those?
 A: Thanks to @Blue for showing me his old answer here which essentially answers my question. The property used from his answer (which is proved in his answer) is as follows:

Property 1. If $P$ is a point on a vertical opening parabola, then the point's horizontal displacement from the vertex is the geometric mean of the parabola's latus rectum and the point's vertical distance from the vertex.

Now here is a diagram from one of @Blue 's answers:

By Property 1. $|KV|^2 = |AK||KC|$ and $|VS|^2 = |AK||KO|$.
So $|KV|^2-|VS|^2 = |AK|(|KC|-|KO|) = |AK||OC|$.
Thus, $(|KV|-|VS|)(|KV|+|VS|) = |AK||OC|$.
But $|KV|-|VS| = OR_{-}$ and $|KV|+|VS| = OR_{+}$.
Which means that $|OR_{-}||OR_{+}| = |AK||OC|$.
For a monic quadratic, $|AK| = 1$ So we get that $|OR_{-}||OR_{+}| = |OC| = c$ as desired.
A: I will use $y = (x - 3)(x - 7)$ to illustrate the idea.
The lines $y = 3$ and $x = 1$ cut the large $7 \times 21$ rectangle into $4$ sub-rectangles.

The diagonal will further cut the rectangle into a $(21 - 3) \times (7 - 6)$ rectangle and a  $(7 - 1) \times (3 - 0)$ rectangle. The two will be equal in area.
Adding the $1 \times 3$ rectangle to both will give the required result.
Added: The sequence of construction is (1) draw the largest rectangle; (2) Let the circle (O, radius = 3) cut the y-axis at (0, 3); (3) draw the line y = 3; (4) Let the ine x = 1 cut y = 3 at (1, 3); (5) join O(0, 0) with P(1, 3) and join P(1, 3) with Q(7, 21). OPQ will then be the diagonal of the largest rectangle.
