Formula for the area of a regular convex pentagon

This question is closely related to my previous question.

Can you provide a proof for the following claim:

In any regular convex pentagon $$ABCDE$$ construct an arbitrary tangent to the incircle of pentagon . Let $$d_1,d_2,d_3,d_4,d_5$$ be a signed distances from vertices $$A,B,C,D,E$$ to tangent line respectively, such that distances to a tangent from points on opposite sides are opposite in sign, while those from points on the same side have the same sign. Denote the side length of pentagon by $$a$$ and the area of pentagon by $$K$$ ,then $$a(d_1+d_2+d_3+d_4+d_5)=2K$$

GeoGebra applet that demonstrates this claim can be found here.

$$\newcommand{real}{\operatorname{Re}}$$Set up a coordinate system where the origin is the pentagon centre; without loss of generality take the pentagon's inradius as $$1$$ and the tangent line as $$x=-1$$. Then $$a=2\tan\frac\pi5$$, the pentagon circumradius $$R=\sec\frac\pi5$$ and $$K=5\tan\frac\pi5$$.
The argument of $$A$$ may be any angle $$\theta$$, but then $$d_1=1+\real(Re^{i\theta})=1+R\real(e^{i\theta})$$. Thus $$d_1+d_2+d_3+d_4+d_5=5+R\real\left(\sum_{k=0}^4e^{i(\theta+2k\pi/5)}\right)$$ The terms in the sum of exponentials are the roots of $$z^5=e^{5i\theta}$$, so by Viète's relations that sum is the negative of the $$z^4$$ coefficient, which is zero. Therefore $$d_1+d_2+d_3+d_4+d_5=5$$ and the equation to prove reduces to $$5a=2K$$, which is easily seen to be true.