Is it possible to derive Reciprocal Pythagoras Theorem from Stewart's theorem with other additional relations?

Question

Consider the following diagram.

The Reciprocal Pythagorean Theorem (RPT) below $$ \frac{1}{t^2} = \frac{1}{b^2}+\frac{1}{c^2} $$ can be easily obtained from \begin{cases} bc=at\\ a^2=b^2+c^2 \end{cases}

I want to get the RPT from Stewart's theorem $$ a(t^2+a_1a_2)=b^2a_2+c^2a_1 $$ and from other additional relation below. \begin{cases} a^2=b^2+c^2\\ a= a_1+a_2 \\ t^2=a_1 a_2 \end{cases}

Question

Is it possible to get RPT from Stewart's theorem plus the additional relations given above?

Answer 1

Simplifying we get $$(b^2+c^2)(a_1^2a_2+a_1a_2^2)-a_1c^4-a_2b^4=0$$