Doubt from Properties of triangles

In triangle ABC, Incentre and circum center are reflections of each other in side BC. Measure angle BAC is

Note that OI is perpendicular to BC since I,B are images of each other across BC. Hence OI bisects BC. It is now easy to see that \triangle ABC is isosceles. So that \angle B = 90^{\circ} -\dfrac{A}{2}

We have \angle OBC = \angle IBC = \dfrac{B}{2}=45^{\circ} -\dfrac{A}{4}

Using properties of circles, \angle BOI = 180^{\circ}-A

Now we have that \angle BOI + \angle OBC = 90^{\circ} so that \angle A = 168^{\circ}

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Correction: \angle A =108^{\circ}

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