An Analytic Geometry approach

The answer given below to the parabola problem , is based on Analytic Geometry and Calculus even it is simple, does not have the charm of pure Geometry where in my opinion Math and Art melt altogether.

So using the same graph posted by G. Markowsky the equality between angles

ADO and BDE implies also the equality between angles ADC and BDC.

Taking under consideration that points B and A belong to a parabola y =X2 and point C is in line connecting the origin with point B m=Yb/Xb we can easily write the coordinates for the above points as follows :

For point B Xb = x ; Yb = x2

For point A Xa = (x – q) ; Yb =(x - q)2

For point C Xc = (x – q) ; Yc = (x – q) Yb/Xb or Yc = x(x – q)

Now tangent of angle BCD = (Yb -Yc)/Xb = (x 2 – x(x – q))/x = q

and tangent of angle CDA = (Yc -Ya)/Xc = (x(x – q) – (x – q)2)/ (x – q) =q

so both angle are equals.

Now let us prove that if angle BCD and CDA are the same point B and A must both lay in a parabola vertex in origin of equation y = Kx2

Since point A and B can be as close as we want we take B a point having the coordinates B(x, y) and for A(x-dx , y-dy) being on the straight line connecting the origin with point B we obtain the coordinates C(x-dx , y-y dx/x )

So tangent CDB = (ydx/x)x and tangent ADC = (-ydx/x+dy)/(x-dx) equalizing both tangents and eliminating denominators we get

y(x – dx) dx = x2 dy – yxdx if we discard (dx)2 we will get the differential equation 2y dx = x dy which render Ln(y)= 2L(x) + Ln(K) or more simple

y = Kx2 So both point A and B belong to the parabola