Let altitude drawn from $C$ meet $AB$ at $D$. Now $\frac{|\triangle APB|}{|\triangle APC|}=\frac{\frac{1}{2}.PD.AB}{\frac{1}{2}.PD.AD}=\frac{AB}{AD}$ We will be done if we prove that $AD$ is rational. In $\triangle ADC$, using Pythagoras theorem we have $AD^2+DC^2=AC^2$ and in $\triangle CDB$ we have $DB^2+DC^2=BC^2$. Subtracting these two equations gives $AD=\frac{AB^2+AC^2-BC^2}{2AB}$ Since $AB, AC, \text{and } BC$ are all integers, $AD$ is rational. Hence Proved