A geometry problem by Matteo Ruth

Let $BA'=x$ , $CB'=y$ and $AC'=z$ . This implies that, $CA'=18-x$ , $AB'=13-y$ and $BC'=10-z$ . Now we find in the figure above that each pair of the right triangles that share a common side are congruent ( ASA ). This leads us to a system of a three variable linear equations:

$10-z=x...(1)$ $18-x=y...(2)$ $13-y=z...(3)$

Solving for $x$ yields that $x=7.5$

No question that Ahmad Saad 's solution is far better that the below solution. But different {long !) approach.
$In\ \Delta\ ABC,\ \ \angle B=Cos^{- 1}\dfrac{10^2+18*2 - 13^2}{2*10*18}=Cos^{-1}\frac{17}{24}...Cos\ Law.\\ r=\dfrac{area\ ABC}{semiperimeter\ ABC}=\dfrac{ \frac 1 4 *\sqrt{41*15*5*21}}{\frac{41}2}=\dfrac{15} 4*\sqrt{\dfrac 7 {41}}.\\ \therefore\ BA'=Cot(\frac 1 2*Cos^{-1}\frac{17}{24})*\dfrac{15} 4*\sqrt{\dfrac 7 {41}}=7.5$ .

We can see that $\color{#D61F06}{AB'=AC'}, \color{#20A900}{B'C=A'C}, \color{#3D99F6}{BC'=A'B}$

$AB=AC'+BC'=10 ...(1) \\ BC=A'B+A'C=18 ...(2) \\ AC=AB'+B'C=13 ...(3)$

• $(1)+(2)-(3):$

$\begin{aligned} (AC'+BC')+(A'B+A'C)-(AB'+B'C) & =10+18-13 \\ (AC'+\color{#3D99F6}{A'B})+(A'B+A'C)-(\color{#D61F06}{AC'}+\color{#20A900}{A'C}) & =15 \\ AC'+A'B+A'B+A'C-AC'-A'C & =15 \\ 2*A'B & =15 \\ A'B & =15/2 \\ A'B & =\boxed{7.5} \end{aligned}$