How do I prove that $\left( 1-\frac{a}{b} \right)\left( 1+\frac{c}{d} \right)=4$? The challenge is to prove $$\left(1-\dfrac{a}{b}\right)\left(1+\dfrac{c}{d}\right)=4.$$

Apart from the Pythagorean theorem, the tangent and secant theorem, and the cosine theorem, I could not invent anything. The calculations here turn out to be very cumbersome and I'm not sure that this is how it should be initially solved.
I had an idea to start from the similarity in the figure, there is a lemma about a right angle on a chord, there is a property of a tangent and a secant...But all this again will lead to cumbersome calculations.

 A: While we wait for an easier method, my answer is based on system of equations using Pythagoras and solving them by hand.

We will use $CE = CF = e, CT = t$ and
$c + d = x$
By Tangent-secant theorem, we have $ \ CF^2 = CT \cdot CD \implies e^2 = t (t+x)$ ...$(i)$
$AC^2 = AH^2 + CH^2 \implies (a+e)^2 =  a^2 - d^2 + (x + t - d)^2$
$e^2 + 2 a e = (x+t)^2 - 2 d (t+x)$
Substituting $e^2$ from $(i), 2ae = x^2 + tx - 2d(t+x)$ ...$(ii)$
Now using similarity of $\triangle ADH$ and $\triangle BDG$, $DG = \dfrac{bd}{a}$
$BC^2 = BG^2 + CG^2$
$\implies (b+e)^2 = b^2 - \left(\dfrac{bd}{a}\right)^2 + \left(t + x + \dfrac{bd}{a}\right)^2$
$2be = x^2 + tx + \dfrac{2bd}{a} (t+x)$ ...$(iii)$
By $(iii) - (ii), (b-a)e = \dfrac{(a+b)d}{a} (t+x)$ ...$(iv)$
By $(ii) \times b + (iii) \times a \ , \ 4abe = x (a + b) (t + x)$
Using $(iv), x = \dfrac{4 b d}{(b-a)}$
As $x = c + d, \ \dfrac{(c+d) (b-a)}{bd} = 4$
$\left(1 - \dfrac{a}{b}\right) \left(1 + \dfrac{c}{d}\right) = 4$
A: It seems like the form of the target relation is trying to tell us something deeper about the configuration that I'm just not seeing. Nevertheless, here's a solution that amounts to a few direct calculations followed by a tedious change of parameters.

With different notation: Let $\triangle ABC$ (with $\angle A\geq \angle B$) have conventional sides $a$, $b$, $c$, which are touched by the incircle at points $D$, $E$, $F$. Let the pairs tangent segments from the vertices have lengths $a'$, $b'$, $c'$. Let cevian $\overline{CF}$ meet the incircle at $U$; and let the perpendicular from $A$ meet that cevian at $V$. Finally, let the perpendicular from $C$ meet side $c$ at $F'$; by Thales' Theorem, $V$ and $F'$ lie on the semicircle with diameter $\overline{AC}$.

Defining
$$t:=|CU| \qquad u:=|UV| \qquad v:=|VF| \qquad f:=|CF|=t+u+v$$
the goal is to show
$$\left(1-\frac{a'}{b'}\right)\left(1+\frac{u}{v}\right)=4 \tag{$\star$}$$
We can write
$$1+\frac{u}{v}=\frac{u+v}{v}=\frac{f-t}{v}=\frac{f^2-ft}{fv} \tag1$$
Multiplying-through by $f$ gives us values we can calculate as follows:
$$\begin{align}
f^2 &= a'^2+b^2-2a'b\cos A &\text{(Law of Cosines)} \tag2\\
ft &= c'^2 &\text{(Sec-Tan Thm wrt incircle)} \tag3\\
fv &= |FA||FF'|=a'(a'-b\cos A)  &\text{(Sec-Sec Thm wrt semicircle $AC$)} \tag4
\end{align}$$
Using the Law of Cosines $$\cos A = \frac{1}{2bc}(-a^2+b^2+c^2) \tag5$$
and rewriting $a=b'+c'$, etc, we find (after a round of symbol-crunching)
$$f^2 - ft = \frac{4a'b'c'}{a'+b'} \qquad 
fv = \frac{a'c'(b'-a')}{a'+b'}\qquad\stackrel{(1)}{\to}\qquad
1+\frac{u}{v}=\frac{4b'}{b'-a'} \tag6$$
from which $(\star)$ follows. $\square$
A: So what I want to say is that I found what could be said to be the "best" way to solve this problem. First, consider the following drawing:

All colored $\triangle$ are pairwise similar to each other, $\left(r_{a} r=a b\right)$.
Yellow $\sim$ blue $\Rightarrow \dfrac{b-a}{2 r}=\dfrac{B_{2} K_{2}}{c+d}$.
Blue $\sim$ red $\Rightarrow \dfrac{B_{1} K_{1}}{d}=\dfrac{2 r}{a}$.
Blue $\sim$ green $\Rightarrow \dfrac{B_{2} K_{2}}{B_{1} K_{1}}=\dfrac{r_{a}}{r}$.
We get:
$$ (b-a)(c+d)=2 B_{2} K_{2} \cdot r=2 B_{1} K_{1} \cdot r_{a}=4 r_{a} r \frac{d}{a}=4 b d $$
