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Show that the line joining the inscribed center to the circumscribed center of a $\triangle ABC$ is inclined to BC at an angle $\tan^{-1}(\frac{\cos B+\cos C-1}{\sin B-\sin C})$

I tried taking the coordinates of $A(0,y_1),B(x_2,0),C(x_3,0)$.So the coordinates of incenter $I$ becomes $\left(\frac{bx_2+cx_3}{a+b+c},\frac{ay_1}{a+b+c}\right)$ and the circumcenter becomes $\left(\frac{x_2+x_3}{2},\frac{y^2_1+x_2x_3}{2y_1}\right)$

But answer is not appearing to come,calculations are messy and clumsy.Please help how should i prove it?

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Consider a $\triangle ABC$ with side BC coinciding with the x-axis such that vertex B is at origin $(0, 0)$ & vertex C is at $(a, 0)$ then vertex A will be at $\left(\frac{a\tan C}{\tan B+\tan C}, \frac{a\tan B\tan C}{\tan B+\tan C}\right)$ Hence, the inscribed center will be $$I\equiv \left(\frac{a\frac{a\tan C}{\tan B+\tan C}+b(0)+c(a)}{a+b+c}, \frac{a\frac{a\tan B\tan C}{\tan B+\tan C}+b(0)+c(0)}{a+b+c} \right)$$ From Sine rule, substitute $a=k\sin A$, $b=k\sin B$ & $c=k\sin C$, we get $$I\equiv \left(\frac{k(\sin A\sin C\cos B+\sin A\sin C)}{\sin A+\sin B+\sin C}, \frac{k\sin A\sin B\sin C}{\sin A+\sin B+\sin C} \right)$$

& the circumscribed center say $D$ can be calculated as $$D\equiv\left(\frac{a}{2}, \frac{a(\tan B\tan C-1)}{2(\tan B+\tan C)} \right)$$ From Sine rule, substitute $a=k\sin A$, we get
$$D\equiv\left(\frac{k\sin A}{2}, \frac{k\cos A\cos B\cos C}{2} \right)$$ Hence, the slope of the line ID joining $I$ & $D$ with the BC (x-axis) is given as $$m=\frac{y_2-y_1}{x_2-x_1}$$

$$=\frac{\frac{k\cos A\cos B\cos C}{2}-\frac{k\sin A\sin B\sin C}{\sin A+\sin B+\sin C}}{\frac{k\sin A}{2}-\frac{k(\sin A\sin C\cos B+\sin A\sin C)}{\sin A+\sin B+\sin C}}$$

By simplifying we get $$m=\frac{\cos B+\cos C-1}{\sin B-\sin C}$$ Hence the angle of the line joining I & D with side BC is given as $$\tan \theta=|m|$$ $$\implies \theta=\tan^{-1}\left|\frac{\cos B+\cos C-1}{\sin B-\sin C}\right|$$

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