$\angle AIO = 90^{\circ} \implies (b-c)(b+c-2a)=0$

Well, here's a hasty solution:

Let $D$ be the intersection of $AI$ with $\odot(ABC)$. Then since $\angle AIO=90^{\circ}$, $I$ must be the midpoint of chord $AD$. Let the length of $ID$ be $d$. By the Incenter-Excenter Lemma, $D$ is the center of the circle containing $\triangle BIC$, so $DB=DC=d$. Then by Ptolemy's Theorem, \[ BC = \frac{AB\cdot DC + AC\cdot DB}{AD} = \frac{2d+3d}{2d} =\frac{5}{2}. \]Thus from Heron's Formula we get \[ [ABC] = \sqrt{\frac{15}{4}\cdot\frac{5}{4}\cdot\frac{3}{4}\cdot\frac{7}{4}} =\frac{15\sqrt{7}}{16}. \]

Ghoshadi wrote: What was ur idea, galav? Denote $M_{AC}$ & $M_{AB}$ as midpoints of $AC$ and $AB$ and let $D_{AC}$ and $D_{AB}$ be touchpoints of the incircle with $AC$ & $AB$ respectively. Since $\angle AIO=90^{\circ}=\angle AM_{AC}O=\angle AM_{AB}O$, we get that $A,O,I,M_{AC},M_{AB}$ lie on circle with $AO$ as a diameter. Further $\angle IAM_{AC}=\angle IAM_{AB}$. So this gives $IM_{AC}=IM_{AB}$ . Moreover $ID_{AC}=ID_{AB}=r$ Also $\angle ID_{AC}M_{AC}=\angle ID_{AB}M_{AB}=90^{\circ}$ . Therefore ,$(D_{AC}M_{AC})^2=(D_{AB}M_{AB})^2 \implies (s-a-\dfrac{b}{2})^2=(s-a-\dfrac{c}{2}) \implies (b-c)(b+c-2a)=0$

Easier yet, see that from shooting lemma, we have since $I$ is the midpt of $AD$ and that the midpoint of $ID$ is on $BC$. Thus $2BC=AB+AC$ and then we have that the area is that $\frac{15\sqrt{7}}{16}$.

Let $M $ and $N $ be the midpoints of $AB $ and $AC $ respectively. Let $AI\cap BC = D $. Trivial angle chasing proves that $\angle AIN $ = $\angle B $. So, $\Delta AIN $ ~ $\Delta ABD $. This implies $\angle IDC $ = $\angle INC $. By congruence of triangles, we get, $CD = CN = \frac {3}{2} $. Similarly, $BD = BM = 1$. Thus, $BC = \frac {5}{2} $. Applying heron's formula, we get, $[\Delta ABC] = \frac {15\sqrt {7}}{16} $.

The solution I submitted: $\textbf{Solution}:$ Let $AI$ intersect the circumcircle at $M$ and the side $\overline{BC}$ at $D$. Note that $M$ is the circumcenter of $\triangle BIC$. Since $\overline{AI}\perp\overline{IO}$, $I$ is the midpoint of $AM$. Also since $\angle MBD=\angle MCB= \angle MAB$, $\triangle MDB \sim \triangle MBA$, which yields that $MB.MD=MB^2=MI^2$\ \ \ \ \ \ \ \-(1). By angle bisector theorem applied to triangles $ABC$ and $BID$, we get that $AI/ID=(b+c)/x=5/x$ where $BC=x$ $\Longrightarrow MA=2MI= 2AI= 10ID/x$ and from (1), $MD=MI^2/MA=MI/2 \Longrightarrow ID= MI/2 \Longrightarrow x=5/2$. Now by an application of Heron's formula, $\Delta= \frac{15\sqrt{7}}{16}$ $\blacksquare$

In∆ABC with AC=10 and BC =15 .The point G and I are it's centroid and incentre .find AB , if angle GIC=90°. ( IFYM ,2012 , BULGARIA ) try this one

Pretty easy. Posting for storage. Let $AI \cap (ABC)=L$, then clearly $I$ is the mid point $AL$ (from the given conditions) Now by Fact-5 it is known that $LB=LC=LI$. We apply ptolemy's theorem $$AC \cdot LB +AB \cdot LC=AL \cdot BC \implies BC=2.5$$Clearly by Heron's we get area as $\frac{15 \sqrt{7}}{16}$