I request your comment about correctness of a parabolic segment construction as given below:
Area of circumscribed chord tangent triangle ABC
By Archimedes thm the area of triangle made by drawing a parallel to chord was shown to be $\frac34 $th.
In the following the area of circumscribed triangle is shown $\frac32 $ times parabolic segment area.
It is shown that yellow area between parabolic arc and chord is $\frac23$ fraction that of triangle $ABC$ between the chord and tangents at ends of chord.
If $(t_2, t_1)$ are slopes at $ (C,A)$ then for a parabola (unit focal length) parameterized as $ (x,y)= (2t, t^2) $ triangle area $ABC$ works out to
$$ ( t_2-t_1)^3/2 $$
The area is evaluated from a determinant of the triangle vertices:
$$ A(2t_1,t_1^2);\; B(2 t_2,t_2^2);\; C(t_1+t_2,t_1t_2);\;$$
The above is found from equation of tangent of the parabola that intersect at $(A,B,C)$ :
$$ y = xt- t^2\,$$
Note that $(x_C,y_C) $ are AM and HM of corresponding coordinates$(x_A,y_A),(x_B,y_B) $;
Area marked yellow is evaluated by integration considering integrand as difference in ordinates
$$\int_{2t_1}^{2 t_2} \dfrac{x^2}{4}-\left(x\;\dfrac{t_1+t_2}{2}- t_1 t_2\right) \;dx = ( t_2-t_1)^3/3 $$
$$ \dfrac{\; Area_{parabolic \;segment}} {\; Area_{chord\;tangents}} =\dfrac23 $$
Has the above result also been established by Archimedes or others later?
Related Construction of internal /external triangles
Since the ratio of circumscribed to inscribed triangles is 2.
Since the areas are in ratios $ 1:2,$ T should be midpoint of MB, sharing the same base, heights are also in the same ratio.AM is mid-point of AC.BTM is median of triangle ABC.
Is my construction correct? Please comment on the same.