Nice. WLOG $b>c$ so points $D,B,C$ lie in that order on one line. $\alpha, \beta,\gamma$ are angle measures associated with vertices $A,B,C$ and $a,b,c$ are segment lengths associated with vertices $A,B,C$ in $\vartriangle ABC$. $DA$ is tangent so $$\angle DAB=\gamma$$$$D\in BC\implies \angle ADB=180^\circ-\beta$$Law of sines in $\vartriangle ABD$ and then in $\vartriangle ABC$ with law of cosines $$|BD|=\frac{c}{\sin(\beta-\gamma)}\cdot\sin \gamma=\frac{c}{\frac{\sin\beta}{\sin \gamma}\cdot\cos\gamma-\cos \beta}=\frac{ac^2}{b^2-c^2}$$$b,c$ are coprime so $$\gcd(b,c)=1\wedge b-c\ge 1$$Assume $|BD|\in Z$. Then $$b^2-c^2|ac^2$$By Euclid's algorithm $$\gcd(b^2-c^2,c^2)=\gcd(b^2,c^2)=\gcd(b,c)^2=1$$Hence $$b^2-c^2|a\implies a\ge |b^2-c^2|=(b+c)\cdot(b-c)\ge b+c>a$$where we used triangle inequality. Above contradiction finishes the proof. Comment As you see it's ok to weaken the assumption from "all side lengths are positive integers which are pairwise coprime" to $\gcd(|AC|,|AB|)=1$ but that could suggest solution a little bit.