A geometry problem by Alisina Zayeni

Let:

$\large \displaystyle \angle{C} = \alpha$

Then:

$\large \displaystyle \angle{B} = 2 \alpha$

$\large \displaystyle \angle{A} = 180^o - 3 \alpha$

Also:

$\large \displaystyle \angle{BAD} = \angle{CAD} = 90^o - \frac{3 \alpha}{2}$

$\large \displaystyle \angle{BDA} = 90^o - \frac{\alpha}{2}$

$\large \displaystyle \angle{CDA} = 90^o + \frac{\alpha}{2}$

Then, by sine law:

$\large \displaystyle \frac{ \overline{AB}} {\sin(\angle{BDA})} = \frac{ \overline{AD}}{\sin(\angle{BAD})}$

$\large \displaystyle \frac{120}{\sin(90^o - \frac{\alpha}{2})} = \frac{50}{\sin(90^o - \frac{3\alpha}{2})}$

$\large \displaystyle \frac{120}{\cos(\frac{\alpha}{2})} = \frac{50}{\cos(\frac{3\alpha}{2})}$

$\large \displaystyle 120 \left [ 4 \cos^3 \left (\frac{\alpha}{2} \right) - 3\cos \left (\frac{\alpha}{2} \right ) \right ] = 50 \cos \left (\frac{\alpha}{2} \right)$

Since $0^o < \alpha < 60^o$ , $\cos \left (\frac{\alpha}{2} \right ) > 0$ . (If $\alpha > 60^o$ , $\angle{B} + \angle{C} > 180^o$ ).

$\color{#20A900} \boxed{ \large \displaystyle \cos \left (\frac{\alpha}{2} \right ) = \sqrt{\frac{41}{48}} }$

Also:

$\large \displaystyle \cos(\alpha) = 2 \cos^2 \left (\frac{\alpha}{2} \right ) - 1 = \frac{17}{24}$

$\large \displaystyle \sin(\alpha) = \sqrt{1 - \cos^2(\alpha)}$

$\color{#20A900} \boxed { \large \displaystyle \sin(\alpha) = \frac{\sqrt{287}}{24} }$

$\large \displaystyle \cos(2 \alpha) = 2 \cos^2 (\alpha) - 1 = \frac{1}{288}$

$\large \displaystyle \sin(2 \alpha) = \sqrt{1 - \cos^2(2 \alpha)}$

$\color{#20A900} \boxed { \large \displaystyle \sin(2 \alpha) = \frac{17 \sqrt{287}}{288} }$

By sine law again:

$\large \displaystyle \frac{ \overline{AC}} {\sin(\angle{B})} = \frac{ \overline{AB}}{\sin(\angle{C})}$

$\large \displaystyle \frac{ \overline{AC}} {\sin(2 \alpha)} = \frac{120} {\sin(\alpha)}$

$\large \displaystyle \overline{AC} = 120 \cdot \frac{ \frac{17 \sqrt{287}}{288}} { \frac{\sqrt{287}}{24} }$

$\color{#3D99F6} \boxed { \large \displaystyle \overline{AC} = 170}$