Let's have some trivialities!

Let $\angle BAC=x$ .

For $BC$ to be the perpendicular biscetor of $OI$ , it follows that the midpoint of $OI$ must be the midpoint of $BC$ as $O$ is on the perpendicular biscetor of $BC$ .

By symettry, $\triangle BIC$ is similar to $\triangle BOC$ by reflecting across $BC$ .

$\angle OBC=\angle IBC=\frac{\angle B}{2}$ because $I$ is on the bisector of $\angle B$ . We have a similar result for $\angle OCB$ .

$\angle BOC=180^{\circ}-\frac{\angle B}{2}-\frac{\angle C}{2}=180^{\circ}-(\frac{\angle B+\angle C}{2})=180^{\circ}-\frac{180^{\circ}-\angle A}{2})=90^{\circ}+\frac{x}{2}$

Let $D$ be a point on the circumcircle of $\triangle ABC$ on the opposite side of $BC$ to $A$ .

$\angle BDC=\frac{90^{\circ}+\frac{x}{2}}{2}=45^{\circ}+\frac{x}{4}$ because angle at circumference is half angle at center.

$ABDC$ is cyclic gives:

$\angle BAC+\angle BDC=180^{\circ} \Rightarrow x+45^{\circ}+\frac{x}{4}=180^{\circ} \Rightarrow \frac{5x}{4}=135^{\circ}$

$x=108^{\circ}$

$\text{Refer to the sketch. I is always within the triangle. But BC is INTERNAL perpendicular }\\ \text{bisector of IO.}\\ \therefore \text{(1) O is outside, (2) ID=DO=r }.~~\implies~(1) ~\angle~A~ >~90^o,~~(2)~d=2r.\\$
link text
$d^2 = R*(R - 2r)=(2r)^2.\\ \implies~4*(\dfrac r R )^2+2*\dfrac r R - 1 =0.\\ Since~\angle~A>90^o,~ we~take ~- tive~sign~and~get~~\dfrac r R= - \dfrac{1+\sqrt5} 4 =CosEOC.\\ \therefore~\angle~EOC~=108^o.\\ \text{BC makes angle 2*EOC at the circumcenter while angle A at the circumference,}\\ \therefore~\angle A=\frac 1 2 *2EOC\\ \implies~\angle~A=108.$

Euler's theorem in geometry

From Wikipedia, the free encyclopedia

[https://en.wikipedia.org/wiki/Euler'stheoremin_geometry)

Let O and I be perpendicular to BC meet BC at D.So in triangle IOD , angle IBD=B/2,ID=r(inradius),BD=a/2(OD perpendicular bisector).So r= $\frac{a}{2}$ tan( $\frac{B}{2}$ ).Also we know r=(s-b)tan( $\frac{B}{2}$ ).Equating we get b=c. Now O is outside and I inside the triangle.angle OBD=angleIOD(since ID=OD given)=B/2.And angleBOD=angleBOA=2C.Alao C=B.Now angle chasing in triangle OBD yields B=B=36.Hence A=108.