There are two possible configurations. Without loss of generality, assume that $B$ lies between $D$ and $C$. Let $a=|B C|, b=|C A|, c=|A B|, x=|B D|$, and $y=|A D|$. By the tangent-secant angle theorem, $\angle B A D=\angle A C B=\angle A C D$, so $\triangle A B D \sim \triangle C A D(\mathrm{AA})$, hence $\frac{|B D|}{|A D|}=\frac{|A B|}{|C A|}=\frac{|A D|}{|C D|}$, or $\frac{x}{y}=\frac{c}{b}=\frac{y}{a+x}$. From this, we get $y c=b x$ and $a c+x c=b y$, so also $b y c=b^{2} x$ and $a c^{2}+x c^{2}=b y c$. Combining these gives $b^{2} x=a c^{2}+x c^{2}$, or $x\left(b^{2}-c^{2}\right)=a c^{2}$.
Now assume for the sake of contradiction that $x$ is an integer. Then $b^{2}-c^{2}$ is a divisor of $a c^{2}$. But we know $\operatorname{gcd}(b, c)=1$, so also $\operatorname{gcd}\left(b^{2}-c^{2}, c\right)=\operatorname{gcd}\left(b^{2}, c\right)=1$. Therefore, $b^{2}-c^{2}$ must be a divisor of $a$. This implies $b^{2}-c^{2} \leq a$. Note that $b^{2}-c^{2}>0$ since $x\left(b^{2}-c^{2}\right)=a c^{2}$; thus, $b-c>0$ as well. Therefore, $b^{2}-c^{2}=(b-c)(b+c) \geq 1 \cdot(b+c)$ because $b$ and $c$ are positive integers. Hence, $a \geq b+c$, which contradicts the triangle inequality. Therefore, $x=|B D|$ cannot be an integer.