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.
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?