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.

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

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    $\begingroup$ Do you have a diagram of the problem? $\endgroup$ – Paul Aug 30 '15 at 10:29
  • $\begingroup$ @Paul,i dont have diagram for this. $\endgroup$ – Brahmagupta Aug 30 '15 at 10:31
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    $\begingroup$ Please try to provide a diagram. And state where your problem originated. Also clarify all the variables, $R$ in particular. $\endgroup$ – MvG Aug 31 '15 at 9:52
  • $\begingroup$ $r_1$ is the radius of escribed circle opposite side $a$ and $R$ is the circumradius. $\endgroup$ – Brahmagupta Aug 31 '15 at 15:37
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    $\begingroup$ Just guessing on the downvoting: No diagram. No identification of initial figure (seems to be a triangle). No clarification of labelling of initial figure. No clarification in the question of quantities in the equation, e.g. $r_1$, $R$.. (Use the "edit" link to the left of the block of questioner data to finish describing the problem.) No attempted solution nor research related to a solution. But that's just a guess. (I neither upvoted nor downvoted.) $\endgroup$ – Eric Towers Sep 9 '15 at 4:56

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

The Law of the sines states that:


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


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