# How show this $\Delta ABC\sim\Delta A'B'C'$

Assmue that:if $\Delta ABC$ and $\Delta A'B'C'$ are not right triangle,and such

$$\sin{(2A)}:\sin{(2A')}=\sin{(2B)}:\sin{(2B')}=\sin{(2C)}:\sin{(2C')}$$

show that $$\Delta ABC\sim\Delta A'B'C'$$

My idea: if $$\sin{A}:\sin{A'}=\sin{B}:\sin{B'}=\sin{C}:\sin{C'}$$ then Law of sines we have $$a:a'=b:b'=c:c'$$ then $$\Delta ABC\sim\Delta A'B'C'$$

But for this ,$$\sin{(2A)}:\sin{(2A')}=\sin{(2B)}:\sin{(2B')}=\sin{(2C)}:\sin{(2C')}$$ I can't prove it ,Thank you

If they are acute triangles, it is simple. Since $\pi-2A, \pi-2B, \pi-2C$ form a triangle, we deduce that $2A=2A'$, etc., hence $A=A'$, etc.
If there is an obtuse angle, we may suppose $A, B< \frac{\pi}{2}$ and $C>\frac{\pi}{2}$, then $A', B'< \frac{\pi}{2}$ and $C'>\frac{\pi}{2}$. Therefore $2A, 2B, 2C-\pi$ and $2A', 2B', 2C'-\pi$, form two triangles. So $2A=2A', 2B=2B', 2C-\pi=2C'-\pi$, the statement follows also.