Find the value

Let the center of the circle be $O$ and its radius $t$ ; and $\angle ABC = \theta$ . Then $\angle ABO = \angle CBO = \frac \theta 2$ . Let $AD=1$ . Then $AD=t$ and $AB=1-t$ . Note that $\tan \frac \theta 2 = t$ . By half-angle tangent substitution , we have

$\begin{aligned} AC & = AB \cdot \tan \theta = (1-t) \cdot \frac {2t}{1-t^2} = \frac {2t}{1+t} \\ AE & = AC\cdot \cos \theta = \frac {2t}{1+t} \cdot \frac {1-t^2}{1+t^2} = \frac {2t(1-t)}{1+t^2} \end{aligned}$

And we have

$\begin{aligned} \frac {\frac 1{AB}+\frac 1{AD}}{\frac 1{AC}+\frac 1{AE}} & = \frac {\frac 1{1-t}+\frac 1t}{\frac {1+t}{2t} + \frac {1+t^2}{2t(1-t)}} = \frac {\frac {1-t+t}{t(1-t)}}{\frac {1-t^2+1+t^2}{2t(1-t)}} = \frac {\frac 1{t(1-t)}}{\frac 1{t(1-t)}} = \boxed 1 \end{aligned}$

Let AB, BC, CA be c, a, b respectively. Then we can find lenght AE using these lenghts (we write area of triangle ABC with lenghts b, c and then with lenghts a, AE): $|AE|=\frac{bc}{a}$ . And also we can write lenght AD using these lenghts: $|AD|=\frac{a+b-c}{2}$ . Now we just put these into the fraction we want to calculate: $\frac{\frac{1}{c}+\frac{2}{a+b-c}}{\frac{1}{b}+\frac{a}{bc}}$ and simplify to this: $\frac{ab+b^2+bc}{a^2-c^2+ab+bc}$ now we can simplify this fruthermore using ( $a^2-c^2=b^2$ ) pythagorean theorem : $\frac{ab+b^2+bc}{b^2+ab+bc}=1$ .

Let radius of circle be $r$ and $2\alpha$ is the angle as shown. These two parameters define this figure.

$AD = r$

$CF = r\text{ tan }\alpha\,$ and $\,AF = r\newline\Rightarrow AC = r(1 + \text{tan }\alpha)$

Also $OF\perp AC\,,\,OG\perp CE\,$ and $\,\angle GOF = 2\alpha \Rightarrow \angle ACE = 2\alpha$

So, $AE = AC\text{ sin}(2\alpha) = r(1 + \text{tan }\alpha)\text{ sin}(2\alpha)$ and $\newline AB = AC\text{ tan}(2\alpha) = r(1 + \text{tan }\alpha)\text{ tan}(2\alpha)$

Writing numerator and denominator separately to avoid complexity in writing.

Numerator :

$\Large\frac{1}{AB}$ $+$ $\Large\frac{1}{AD}$ $=$ $\Large\frac{1}{r(1 + \text{tan }\alpha)\text{ tan}(2\alpha)}$ $+$ $\Large\frac{1}{r}$

$\hspace{45pt} = \Large\frac{1}{r(1 + \text{tan }\alpha)}\Bigg(\frac{1}{\text{tan}(2\alpha)}$ $+$ $\text{tan }\alpha + 1\Bigg)$

$\hspace{45pt} = \Large\frac{1}{r(1 + t)}\Bigg(\frac{1 - t^2}{2t}$ $+\, t + 1\Bigg)\hspace{50pt} [t = \text{tan }\alpha]$

$\hspace{45pt} = \Large\frac{1}{r(1 + t)}\Bigg(\frac{t^2 + 2t + 1}{2t}\Bigg)$ $=$ $\Large\frac{t+1}{2rt}$

Denominator :

$\Large\frac{1}{AC}$ $+$ $\Large\frac{1}{AE}$ $=$ $\Large\frac{1}{r(1 + \text{tan }\alpha)}$ $+$ $\Large\frac{1}{r(1 + \text{tan }\alpha)\text{ sin}(2\alpha)}$

$\hspace{45pt} = \Large\frac{1}{r(1 + t)}$ $\Bigg(1 + \Large\frac{1 + t^2}{2t}\Bigg)$

$\hspace{45pt} = \Large\frac{1}{r(1 + t)}$ $\Bigg(\Large\frac{t^2 + 2t + 1}{2t}\Bigg)$

$\hspace{45pt} = \Large\frac{t+1}{2rt}$

We get Numerator $=$ Denominator which means $\Large\frac{\text{Numerator}}{\text{Denominator}}$ $= 1$