Triangle with $q$ identical lines

The task is: Given a triangle with $q$ identical lines, in this way as you can see on the picture. The orange lines are the same. Calculate $\gamma$!

$\Delta ABC'$ is an isosceles triangle, therefore $\beta_1=\gamma$.

The sum of the angles at the B point is $180^{\circ}$, because AH is a line: $\cancel{180^{\circ}}-2\gamma+\beta_2=\cancel{180^{\circ}}$ $\beta_2=2\gamma$

From the isosceles triangles and the sum of angles at the C point: $\beta_1+\cancel{180^{\circ}}-2\beta_2+\beta_3=\cancel{180^{\circ}}\implies \beta_3=4\gamma-\gamma =3\gamma$ Same way $\beta_4=4\gamma, \beta_5=5\gamma, \cdots, \beta_n=n\gamma$, because if we see an angle: $\beta_{n-2}+\cancel{180^{\circ}}-2\beta_{n-1}+\beta_n=\cancel{180^{\circ}}\implies \beta_n=2\beta_{n-1}-\beta_{n-2}=(2(n-1)-n+2)\gamma=n\gamma$ This is true for the first, second and third angles, so this must be true for every angles.

If we have a triangle with $q$ lines, then the last $\beta$ is the $\frac{q-3}{2}$th.

Now in the bottom $HG'H'$ triangle: $\alpha+\alpha+\alpha+\beta_6+\beta_6=180^{\circ}$ or generally $3\alpha+2\beta_{\frac{q-3}{2}}=180^{\circ}$. But if we see the big triangle $2\alpha+2\beta_{\frac{q-3}{2}}+\gamma=180^{\circ}$. Therefore $\alpha=\gamma$. After substituting: $3\gamma+2\beta_{\frac{q-3}{2}}=180^{\circ}$. Replacing $\beta_{\frac{q-3}{2}}$: $3\gamma+2\cfrac{q-3}{2}\gamma=180^{\circ}$ $3\gamma+q\gamma-3\gamma=180^{\circ}$ $q\gamma=180^{\circ}$ $\boxed{\gamma=\cfrac{180^{\circ}}{q}}$

Now can you solve this?