Yeah, with the law of sines, one can get such right angles, but this is basically by constructing a new set of triangles. See https://stacker.news/items/741908/r/south_korea_ln?commentId=743146

What makes this so tricky is that there is no right triangle involved, yet you look for one because $$sin()$$ is involved. At least I did for too long. Thank you!

Scroogey

I guess the Ptolemy hint did the trick. 
Or maybe with just the figure, you'd have figured it out too...

There are some beautiful other problems in Math Horizons, I'll have to take some time to curate the ones that are easy to post here.

south_korea_ln

science

[Daily puzzle] Proving a curious equality

![cirle](https://i.ibb.co/V9xsnSL/circle.png)

Angle units are $$\frac{1}{7}\pi$$.

Angle sums in triangles must be 7 (=180°), in quadrilateral 14 (=360°).

Note two pairs of similar triangles. One pair is isosceles.

Ptolemy says

$$
c_1*c_2 + c_1*c_3 = c_2*c_3
$$

Hence

$$
sin(\frac{1}{7}\pi)*sin(\frac{2}{7}\pi) + sin(\frac{1}{7}\pi)*sin(\frac{3}{7}\pi) = sin(\frac{2}{7}\pi)*sin(\frac{3}{7}\pi)  
$$

Equals

$$
sin(\frac{1}{7}\pi) = \frac{sin(\frac{2}{7}\pi)*sin(\frac{3}{7}\pi)
}{sin(\frac{2}{7}\pi)+sin(\frac{3}{7}\pi)}
$$

Equals

$$
\frac{1}{sin(\frac{1}{7}\pi)} = \frac{1}{sin(\frac{2}{7}\pi)} + \frac{1}{sin(\frac{3}{7}\pi)}
$$
