If $AD=BD$, $\angle ADC=3\angle CAB$, $AB=\sqrt{2}$, $BC=\sqrt{17}$, $CD=\sqrt{10}$. Find $AC$ In quadrilateral $ABCD$, we have
$$AD=BD,\angle ADC=3\angle CAB,AB=\sqrt{2},$$ $$BC=\sqrt{17},CD=\sqrt{10}$$
Find the $AC=?$
My idea: let $$\angle CAB=x.\angle ADC=3x,\angle ADB=y,$$
then we have
$$\angle CAD=90-\dfrac{y}{2}-x,\angle ACD=\dfrac{y}{2}+90-2x$$then we have
$$\dfrac{\dfrac{\sqrt{2}}{2}}{\sin{\dfrac{y}{2}}}=BD$$
and
$$\dfrac{BD}{\sin{BCD}}=\dfrac{DC}{\sin{DBC}}=\dfrac{BC}{\sin{BDC}}$$
then
$$\dfrac{\sqrt{2}}{\sin{\dfrac{y}{2}}\sin{BCA}}=\dfrac{\sqrt{10}}{\sin{DBC}}=\dfrac{\sqrt{17}}{\sin{(3x-y)}}$$
and in $\Delta ABC$,we have
$$\dfrac{\sqrt{17}}{\sin{x}}=\dfrac{\sqrt{2}}{\sin{ACB}}=\dfrac{AC}{\sin{ABC}}$$
in $\Delta ADC$,we have
$$\dfrac{AC}{\sin{3x}}=\dfrac{\sqrt{10}}{\sin{(90-y/2-x)}}=\dfrac{AD}{\sin{(90+y/2-2x)}}$$then I fell very ugly,and   I can't.maybe have other idea. Thank you
 A: $ABCD$ on a square lattice">
With three distances given, each as the square root of an integer which is the sum of two squares, this problem looks as if the points are meant to have integer $x$ and $y$ coordinates. So treat the problem as one in coordinate geometry.
We are given:
\begin{align}
AD&=BD\tag{1},\\
\angle ADC&=3\angle CAB\tag{2},\\
AB&=\sqrt2\tag{3},\\
BC&=\sqrt{17}\tag{4},\\
CD&=\sqrt{10}.\tag{5}
\end{align}
To satisfy (3), define cartesian coordinates so that $A=(0, 1)$ and $B=(1, 0)$. By (1),
$D$ lies on the line $l$ with equation $y=x$.
Next, locate $C$ by trying each lattice point satisfying (4),
then seeing if there is a suitable location for $D=(x, x)$ satisfying (5).
If $C=(0, 4)$ or $(2, 4)$ then $\angle CAB>90^\circ$ so (2) cannot be satisfied with the angles' values correctly specified.
If $C=(5, -1)$ then no $D$ on $l$ is near enough $C$.
If $C=(5, 1)$ then $D=(2, 2)$ or $(4, 4)$. $\angle CAB=45^\circ$ so $\angle ADC=135^\circ$ by (2). $D=(4, 4)$ makes $\angle ADC$ too small. However, $D=(2, 2)$ makes $\angle ADC=135^\circ$ as required, because, where $E=(3, 0)$, $\angle ADE=90^\circ$ and $\angle EDC=45^\circ$ as it is one of the acute angles of an isosceles right-angled triangle. In this case $AC=5$.
$C=(0, -4)$ and $C=(2, -4)$ yield reflections in $AB$ of the situation when $C=(5, 1)$ and $C=(5, -1)$ respectively, so these need not be further checked.
Thus $AC=5$.
A: You only have to solve the 4-equation system below:
\begin{cases}
\overline{AC}^2+\overline{AB}^2-2\overline{AC}\cdot \overline{AB}\cos(x)=\overline{BC}^2 \\
\overline{AD}^2+\overline{CD}^2-2\overline{AD}\cdot \overline{CD}\cos(3x)=\overline{AC}^2 \\
\overline{BD}^2+\overline{CD}^2-2\overline{BD}\cdot \overline{CD}\cos(3x-y)=\overline{BC}^2 \\
\frac{\overline{AD}}{\sin\left(\frac{\pi-y}{2}\right)} = \frac{\overline{AB}}{\sin(y)} \\
\end{cases}
The first 3 equations can be obtained by applying Carnot Theorem in the triangles $\Delta ABC$, $\Delta ACD$ and $\Delta BDC$, while the last is easy to find by applying the Law of Sines in $\Delta ABD$. However there are only 4 variables ($\overline{AC}$,$\overline{AD}$, $x$ and $y$) beacuse $\overline{AD} = \overline{BD}$, so the system becomes easier to solve:
\begin{cases}
\overline{AC}^2+2-2\sqrt{2}\cdot \overline{AC}\cos(x)=17 \\
\overline{AD}^2+10-2\overline{AD}\cdot \sqrt{10}\cos(3x)=\overline{AC}^2 \\
\overline{AD}^2+10-2\overline{AD}\cdot \sqrt{10}\cos(3x-y)=17 \\
\frac{\overline{AD}}{\cos\left(\frac{y}{2}\right)} = \frac{\sqrt{2}}{\sin(y)} \\
\end{cases}
