A clue for the geometry problem

Question

With respect to the picture $AB=DC , \ B=10\alpha ,A_1=\alpha , c=2\alpha$ so need $\alpha$ I tried first to $A=A_1+A_2=\alpha +\beta$ so $$\beta =180-13\alpha$$ then draw a line $AF=AB$ as below or $DG=AB=DC$

but got nothing, (seems like parallel lines, but I have no evidence to show this!) .

I get stuck on this after about 2 hours. Actually, I was far from geometry for years. Now I want a clue or a hint to take over and find a solution. Thanks in advance
Remark: I tried to write $sine/cosine $ theorem ,but needs the relation for $5x $ arcs and become more complicated.

Answer 1

You can also use this figure. We have:

$AD_1||BC\Rightarrow \overset{\large\frown}{AB}=\overset{\large\frown}{D_1C}=4\alpha$

$\overset{\large\frown}{D_1I}=\frac 12 \overset{\large\frown}{D_1C}=2\alpha$

$\overset{\large\frown}{CF}=2\alpha=\overset{\large\frown}{IC}$

$\overset{\large\frown}{AD_1C}=20\alpha$

$\overset{\large\frown}{AD_1C}-\overset{\large\frown}{IC}=20\alpha-2\alpha=18\alpha=180^o$

That is $\alpha=10^o$

Answer 2

Hint: connect $B$ and $G$. $\triangle ABG$ is isosceles. Use this fact to write an equation involving $\measuredangle ABG$ (which is equal to $\measuredangle AGB$), and show that $\measuredangle ABG =\measuredangle AGB = 6\alpha$.

We have $$\measuredangle BGD = 180^\circ - 2\alpha - \measuredangle AGB.$$ Thus, using the sum of the interior angles of $\triangle BGD$ we get \begin{eqnarray}\measuredangle BGD &=& 180^\circ - 4\alpha - (180^\circ-2\alpha-\measuredangle AGB)=\\&=&\measuredangle AGB-2\alpha.\end{eqnarray} Since $\measuredangle ABG + \measuredangle GBD = 10\alpha$, and $\measuredangle AGB = \measuredangle ABG$, we have $$2\measuredangle AGB - 2\alpha = 10\alpha.$$

So there is another isosceles triangle in the Figure. Which one? Can you reach your result from here?

We get $\measuredangle GBD = \measuredangle GDB = 4\alpha$. Hence $BG \cong GD$. Therefore $\triangle ABG$ is equilateral and $\alpha = 10^\circ.$

Answer 3

(A nice another way) In your figure, one has in triangle $\triangle{ACB}$ $$\frac{AC}{\sin(10\alpha)}=\frac{AB}{\sin(2\alpha)}$$ and in triangle $\triangle{ACD}$ $$\frac{AC}{\sin(180-3\alpha)}=\frac{DC}{\sin(\alpha}$$ from which the equation in $\alpha$ $$\sin(10\alpha)=\frac{\sin(2\alpha)\sin(3\alpha)}{\sin(\alpha)}$$ By periodicity, there are many solutions but we have the restriction $0\lt 13\alpha\lt180^{\circ}$ so $0\lt\alpha\lt13.8462^{\circ}$. Consequently we choose the smallest positive solution which is $\boxed{\alpha=10^{\circ}}$ and the angles of the triangles in question are $100, 20$ and $60$ degrees.