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Show that the line joining the incenter ...

Show that the line joining the incenter to the circumcentre of triangle ABC is inclined to the side BC at an angle `tan^(-1) ((cos B + cos C -1)/(sin C - sin B))`

Text Solution

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Let `I` be the incenter and O be the circumcenter of the triangle ABC.

Let OL be parallel to BC
Let `angle IOL = theta`
Now, `IM = r, OC = R, angle NOC = A`. Then
`tan theta = (IL)/(OL) = (IM- LM)/(BM- BN)`
`= (IM - ON)/(BM - NC)`
`= (r - R cos A)/(r cot. (B)/(2) - R sin A)`
`=(4R sin.(A)/(2) sin.(B)/(2) sin.(C)/(2) - R cosA)/(4R sin.(A)/(2) sin.(B)/(2) sin.(C)/(2). cot. (B)/(2)- R sin A)`
`=(cos A + cos B + cos C - 1- cos A)/(sin A + sin C - sin B - sin A)`
`= (cos B + cos C -1)/(sin C - sin B)`
or `theta = tan^(-1) [(cos B + cos C -1)/(sin C - sin B)]`
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