Prove that the height of this triangle equals the sum of the radii of the cirles

Question

Take any equilateral triangle $ABD$, then extend the side $AB$ into a line. Take a point $C\in \overleftrightarrow{AB}$, and trace the line $\overleftrightarrow{CD}$. Circles $G$ and $E$ are both tangent to the equilateral triangle, and to the lines $\overleftrightarrow{AB}$ and $\overleftrightarrow{AD}$. Prove that the height of the triangle $ABD$ equals the sume of the radii of the circles $G$ and $E$, regardless the position of point $C$ in the line $\overleftrightarrow{AB}$. Here's an image with some other elements:

Note: Dotted lines in the image represent angle bisectors of the angles they cut through. It's easy to show that the centers of the circles must lie on those angle bisectors. $M$ is the mid-point of side $AB$. Some tangency points, such as $I$ and $J$ have also been marked.

Answer 1

$$\left.\begin{align} p+r &= 2t \\ q+s &= 2t \\ p+q &= 2t + r + s \end{align} \quad \right\rbrace \implies \quad t = r + s$$

Answer 2

\begin{align} r&=|GI| ,\quad r_a=|EF| ,\quad h=|DM|=r_a+|EL| ,\\ \angle ADC&= \tfrac\pi3-2\phi =2\psi ,\\ \angle EDB&=\tfrac\pi6+\phi ,\\ \angle LDE&=\tfrac\pi6-\phi =\tfrac12\angle ADC =\psi ,\\ \angle EKD&=\angle GAD=\tfrac\pi3 ,\\ |DK|&=|DA| ,\\ \triangle ADG &\cong \triangle KDE ,\\ |EL|&=|GI|=r . \end{align}