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Let H be the orthocenter of triangle ABC, then angle subtended by side BC at the centre of incircle of triangle CHB is

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$\frac{A}{2} + \frac{π}{2}$

$\frac{B + C}{2} + \frac{π}{2}$

$\frac{B - C}{2} + \frac{π}{2}$

$\frac{B + C}{2} + \frac{π}{4}$

answer is B.

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Detailed Solution

$\begin{aligned} ∠ B I C = \frac{π}{2} + (\frac{π - A}{2}) \\ = \frac{π}{2} + (\frac{B + C}{2}) \end{aligned}$

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