Find two ratios given an inequality about a triangle with integer coordinates

Question

Let $A(0,0)$, $B(p,q)$, $C(r,s)$ where $p,q,r,s \in \mathbb{Z}$. Also, it is known that $(|AB|+|BC|)^2<8\cdot \textbf{Area}(\triangle{ABC})+1$

$1.$ If $\triangle{ABC}$ is reflected across side $AC$, so that we get the point $D$. Then the quadrilateral $ABCD$ is formed where $\frac{AC}{BD}$ is equal to $n$. Find $n$

$2.$ In the quadrilateral $ABCD$, if $\angle{ABC}=k\cdot\angle{DAC}$ then find $k$

I drew the diagram but I don't know in particular what does reflected across side AC means$?$ I have encountered problems of reflections where we have to reflect the whole triangle on $y$ or $x$ axis or origin, but this is new to me. Nevertheless, I assumed that it means a point $D$ such that $\triangle{ABC}\cong\triangle{ADC}$. Moving ahead, I can't really find the ratio asked. Everything is coming in variables and I'm not able to co-relate them.

For the second one, I applied sine rule but in vain. Any help is appreciated.

EDIT

According to the answer given in my book, $n=1$ and $k=2$.

Answer 1

This problem is, in fact, about inequalities and the discreteness of integers besides the basic knowledge of geometry.

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$$\begin{aligned}&\quad\quad1\\ &>(|AB|+|BC|)^2-8\cdot\text{Area}(\triangle{ABC})\\ &=(|AB|^2+|BC|^2-4\cdot\text{Area}(\triangle{ABC}))+2( |AB|\cdot|BC|-2\cdot\text{Area}(\triangle{ABC}))\\ &\ge|AB|^2+|BC|^2-4\cdot\text{Area}(\triangle{ABC})\\ &=(|AB|-|BC|)^2+2(|AB|\cdot|BC|-2\cdot\text{Area}(\triangle{ABC}))\\ &\ge0. \end{aligned}$$

$|AB|^2=p^2+q^2$ is an integer.
$|BC|^2=(r-p)^2+(s-q)^2$ is an integer.
Since, as illustrated below, $\triangle{ABC}$ is the rectangle minus three (possibly degenerate) triangles, where all coordinates of all vertices are integers, $2\cdot\text{Area}(\triangle{ABC})$ is an integer.

Hence, $|AB|^2+|BC|^2-4\cdot\text{Area}(\triangle{ABC})$ is an integer. Since it is $<1$ and $\ge0$, it must be $0$. That is,

$$(|AB|-|BC|)^2+2(|AB|\cdot|BC|-2\cdot\text{Area}(\triangle{ABC}))=0.$$ Since $(|AB|-|BC|)^2\ge0$ and $|AB|\cdot|BC|-2\cdot\text{Area}(\triangle{ABC})\ge0$, we must have $(|AB|-|BC|)^2=0$ and $|AB|\cdot|BC|-2\cdot\text{Area}(\triangle{ABC})=0$. The former implies $AB$ and $BC$ are equal; the latter implies $AB$ and $BC$ are perpendicular to each other.

Hence, $ABCD$ is a square.

1. $\frac{AC}{BD}=1$
2. $\angle{ABC}=90^\circ$, $\angle{DAC}=45^\circ$. $k=2$.