# if $AD=BD,\angle ADC=3\angle CAB,AB=\sqrt{2},BC=\sqrt{17},CD=\sqrt{10}$,How find $AC$

in quadrilateral $ABCD$,such $$AD=BD,\angle ADC=3\angle CAB,AB=\sqrt{2},BC=\sqrt{17},CD=\sqrt{10}$$ 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

• You mentioned that ABCD is a parallelogram but your drawing of it looks like an ordinary quadrilateral. Besides, If it is a parallelogram, then the opposite sides should be equal, but clearly AB =/= CD. – Mick Apr 7 '14 at 15:41

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}