Prove that $\sin\theta_1.\sin\theta_2.\sin\theta_3=\frac{r^2_1}{16R^2}$

Question

If $2\theta_1,2\theta_2,2\theta_3$ are the angles subtended by the circle escribed to the side $a$(opposite to vertex $A$) of a triangle at the centers of the inscribed triangle and the other two escribed circles,prove that $\sin\theta_1.\sin\theta_2.\sin\theta_3=\frac{r^2_1}{16R^2}$

Here $r_1$ is the radius of the circle escribed to the side $a$(opposite to vertex $A$) and $R$ is the radius of the circumcircle.

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I dont know what is the relation satisfied by $\theta_1,\theta_2,\theta_3$ and could not think more about it.Please help me in solving this question

Answer 1

I am trying to reverse engineer the solution to the question.

The Law of the sines states that:

$$\frac{a}{\sin{\alpha}}=\frac{b}{\sin{\beta}}=\frac{c}{\sin{\gamma}}=2R$$

Thus, In a triangle: $$\frac{\sin{\alpha}}{a}=\frac{\sin{\beta}}{b}=\frac{\sin{\gamma}}{c}=\frac{1}{2R}$$

In your case, I assume we are talking about a Quadrilateral with sides of size $a$,$b$,$c$ and $d$ and with angles $\theta_1$,$\theta_2$,$\theta_3$ and $\theta_4$.

And thus: $$\frac{\sin{\theta_1}}{a}\frac{\sin{\theta_2}}{b}\frac{\sin{\theta_3}}{c}\frac{\sin{\theta_4}}{d}=\frac{1}{16R^4}$$

$$\sin{\theta_1}\sin{\theta_2}\sin{\theta_3}=\frac{abcd}{\sin{\theta_4}16R^4}$$ And all we have left to understand / explain, is why $$\frac{abcd}{\sin{\theta_4}}=r_1^2$$

Again, a diagram would help