trapezium and triangle sum conjecture

let there exist an isosceles triangle t with side length x and height h. a trapezium T may be constructed within the triangle with a major base x, a minor base a, and height h₁. a formula may be derived to get a height h_n to make a trapezium within t that occupies an m factor of the area A_t.

$\cot\theta=\frac{h_1}{\frac{x}{n}}$
$\therefore\;\frac{x\cot\theta}{n}=h_1$
$\underline{\mathrm{let:}}\;a_1+a_2=\frac{x}{2}$
$\tan\theta=\frac{a_1}{\frac{h_1}{2}}$
$\therefore\;\frac{h_1\tan\theta}{2}=a_1$
$\therefore\;\frac{x\cot\theta\tan\theta}{n^2}=a_1$
$\therefore\;\frac{x}{n^2}=a_1$
$\cot\theta=\frac{a_2}{h_1}$
$\therefore\;h_1\cot\theta=a_2$
$\therefore\;\frac{x\cot\theta\cot\theta}{n}=a_2$
$\therefore\;\frac{x\cot^2\theta}{n}=a_2$
$\underline{\mathrm{let:}}\;A_1=h_1a_1$
$\therefore\;A_1=\frac{x\cot\theta}{n}(\frac{x}{n^2})$
$\therefore\;A_1=\frac{x^2\cot\theta}{n^3}
$\underline{\mathrm{let:}}\;A_2=\frac{h_1a_2}{2}$
$\therefore\;A_2=\frac{{(\frac{x\cot\theta}{n})(\frac{x\cot^2\theta}{n})}}{2}
$\therefore\;A_2=\frac{x^2\cot^3\theta}{2n^2}$
$A_T=2A_1+2A_2$
$A_T=2(\frac{x^2\cot\theta}{n^3})+2(\frac{x^2\cot^3\theta}{2n^2})$
$A_T=x^2(\frac{2\cot\theta}{n^3}+\frac{\cot^3\theta}{n^2})$
$A_T=x^2(\frac{2\cot\theta}{n^3}+\frac{n\cot^3\theta}{n^3})$
$A_T=x^2(\frac{2\cot\theta+n\cot^3\theta}{n^3})$
$A_T=\frac{x^2(2\cot\theta+n\cot^3\theta)}{n^3}$
$A_t=\displaystyle\left[\sum_{n=1}^{m-1}nA_T\right]+\displaystyle\frac{x^2\sec\theta}{m}
$\therefore\;\underline{A_t=\displaystyle\left[\sum_{n=1}^{m-1}\frac{x^2(2\cot\theta+n\cot^3\theta)}{n^2}\right]+\displaystyle\frac{x^2\sec\theta}{m}}$
$\Box$