Trigonometry! #9

I will try , but I did it in other way ,

Take O coordinates as (0,0) and other as (a,b) , now by quadrants the coordinates signs will change and then applying mid point formula we can find the coordinates of the vertices of the triangle , and we can see that each side length is equal so equilateral or we can say that all the three coincide at one point , thus equilateral

Vectors and complex numbers would often offer easier bookkeeping, especially since there is so much rotation around.

Let the points be on the unit circle, with $A = \alpha, C = \beta , E = \gamma$ . Since $AB=CD=EF = r$ , we can treat $B, D, F$ as rotation of $A, C, E$ by the $60 ^ \circ$ = $\omega$ , where $\omega ^ 6 = 1$ . Thus $B = \alpha \omega, D = \beta \omega, F = \gamma \omega$ .

Then, $I$ is the midpoint of $BC$ , so its represented by $\frac{ \alpha \omega + \beta } { 2}$ . Likewise, $K = \frac{\gamma \omega + \alpha } { 2}$ and $J = \frac{ \beta \omega + \gamma } { 2}$ .

Now, we get that $\vec{KI} = \vec{I} - \vec{K} = \frac{ \alpha ( \omega - 1 ) + \beta - \gamma \omega } { 2}$ and likewise $\vec {IJ } = \frac{ - \alpha \omega + \beta ( \omega - 1) + \gamma } { 2}$ .

We can now easily verify that $\omega ^2 \vec{KI} = \vec{ IJ }$ which tells us that they are equal length, and meet at an angle of $2 \times 60 ^ \circ$ . Thus, we have an equilateral triangle.