Prove that $\angle AIO = 90^{\circ}$ with $I$ being the incenter and $O$ circumcenter of $\triangle ABC$

Question

let $O_1$ be the center of the semicircle and $XY$ be its diameter. We must have $O_1A \perp XY$ and $O_1A > 2R = XY$ otherwise $\triangle ABC$ doesn't exist.

I also know that the semicircle is the $A$ excircle of $\triangle ABC$ and I also happen to know that $\angle AIO = 90^{\circ}\iff a = \frac{b+c}2$ where $a,b,c$ are the sides of $\triangle ABC$.

I found this problem to be quite challenging still, because it is hard to work with the constrain of a reflection over a line.

EDIT. all the details: we are given a circle $\omega = (O_1,R)$ and a point $A$ such that $O_1A > 2R$. $XY$ is the diameter of $\omega$ perpendicular to $O_1A$. Line $A_1A_2$ is the polar line of $A$ w.r.t. $\omega$. We reflect line $XY$ over line $A_1A_2$ and obtain the dotted line parallel to $XY$ which cuts $\omega$ at $T$. $B$ is the meeting of $AA_2$ with the tangent line from $T$ and $C$ is the meeting of that tangent line with $AA_1$. I want to prove that $\angle AIO = 90^{\circ}$ with $I$ and $O$ being respectively the incenter and circumcenter of $\triangle ABC$

EDIT: since all of you guys were so excited to prove my little $2a=b+c$ lemma, here is a little reward for you:

Answer 1

I am trying to give a clean synthetic approach. It restates the given constellation having $\Delta ABC$ in foreground. (And not starting with the irrelevant diameter $XY$.) Then we translate all points and properties in terms of this triangle. For instance, $O_1$ is the excenter $I_A$. And the condition on $T$ gets translated as "mid point of $AI_A$ is on $BC$". Known metric relations in a triangle, and formulas for the placement of the incenter, and excenters together with their projections on the sides are used. The answer became longer then it was initially intended, as i tried to introduce notations, restate clearly the bountied proposition by using them, collect facts, and make them conclude in a short computation. Let's start.

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Observe first that $CO_1$ is the angle bisector of $\widehat{TCA_1}$, since $CT$ and $CA_1$ are tangents from $C$ to a circle centered in $O_1$. (The triangles $\Delta CTO_1$ and $\Delta CA_1O_1$ are congruent, one right angle each, one radius side each, and a common side.) Similarly, $BO_1$ bisects $\widehat{TBA_2}$, so $O_1$ is the $A$-excenter of $\Delta ABC$. We would like to state the given constellation of points by starting from $\Delta ABC$ in an equivalent manner. The one needed translation is that of the condition that $T$ projects on $O_1A$ in the reflection of $O_1$ w.r.t. $A_1A_2$. So we...

Claim: Let $A$ be a point, which is exterior to a circle $\omega$ centered in a point $O_1$ with radius $R$. Denote by $*$ the inversion $X\to X^*$ in $\omega$. We draw the tangents from $A$ to the circle, let $A_1$, $A_2$ be the points of tangency. In particular, $A_1A_2$ is the polar of the pole $A$ with respect to $\omega$. And $A^*$ is $A_1A_2\cap O_1A$. Let $S$ be the reflection of $O_1$ w.r.t. the polar $A_1A_2$.

Let now $T$ be a point on the arc $\overset\frown{A_1A_2}$ (which is in the half-plane w.r.t. $A_1A_2$ containing $A$). Draw the tangent in $T$ to $\omega$. It intersects $AA_1$, and $AA_2$ in $C$, respectively $B$. Let $M$ be the intersection of lines $$ M=BTC\cap O_1A\ . $$ Then $T$ projects in $S$ on $O_1A$, if and only if (iff) $M$ is the mid point of $AO_1$.

Proof: Assume that $T$ projects in $S$. Then in the triangle $\Delta TMO_1$ with a right angle in $T$ we have a formula for $O_1S\cdot O_1M$ which leads to: $$ O_1A^*\cdot O_1A =O_1A_1^2=R^2=O_1T^2=O_1S\cdot O_1M=2O_1A^*\cdot O_1M\ . $$ So $M$ is the mid point of $O_1A$. For the converse it is easy to adapt the same argument. (Denote by $S'$ the projection of $T$ on $O_1M$, and then similarly $O_1A^*\cdot O_1A =O_1S'\cdot O_1M=O_1S'\cdot 2O_1M$, so $S'=S$.)

$\square$

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Using the above, we can then restate the given problem:

Lemma: Let $\Delta ABC$ be a triangle with incenter $I$ and $A$-excenter $I_A$. Its sides are $a,b,c$. Assume that the mid point $M$ of $AI_A$ lies on $BC$. Then: $2a=b+c$.

Proof: Denote by $S$ the area of $\Delta ABC$, by $s$ the half-perimeter $s=\frac 12(a+b+c)$, by $r$, $r_A$ the in-radius and the $A$-ex-radius, by $h_A$ the height from $A$. Then: $$ \bbox[yellow]{\qquad r_A(s-a)=S=\frac 12ah_A\ .\qquad} $$ (The formula $S=r_A(s-a)$ is similar to the handy known formula $S=rs$, shown by area splitting $\Delta ABC$ in $\Delta IBC$, $\Delta ICA$, $\Delta IAB$. When replacing $I$ bay $I_A$, the first area is taken with the negative sign in the split.)

The assumption on $M$ shows that $A$ and $I_A$ have the same distance to $BC$: $$ \bbox[yellow]{\qquad h_A=r_A\ .\qquad} $$ This gives $s-a=\frac a2$, so $a+b+c=2s=3a$, so $b+c=2a$.

$\square$

Observation: By the same argument, the converse is also true.

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It remains to show:

Proposition: Let $\Delta ABC$ be a triangle. We use same notations as in the above lemma. Let us also introduce the circumcenter $O$. Assume that $$2a=b+c\ ,$$ and that $O\ne I$. Then $AI\perp OI$.

Proof: We denote by $R$ the circumradius, and use Euler's relation $OI^2=R^2-2Rr$, and Heron's formula.

We are successively using below equivalent steps from one line to the next one to translate $AI\perp OI$ into a relation between $a,b,c$. $$ \begin{aligned} AI&\perp OI\ ,\\ AI^2 &=AO^2-OI^2\ ,\\ AI^2&= R^2-(R^2-2Rr) \ ,\\ AI^2 &= 2Rr\ ,\\ \frac r{2R} &=\frac{r^2}{AI^2}\ .\\[3mm] &\quad\text{Separated computation:}\\ &\quad\frac{r^2}{AI^2}=\sin ^2\frac A2=1-\cos ^2\frac A2 =\frac {1-\cos A}2 =\frac 12\left(1-\frac{b^2+c^2-a^2}{2bc}\right)\\ &\quad\qquad=\frac 1{4bc}(a^2-(b-c)^2)=\frac 1{4bc}(a-b+c)(a+b-c) \\ &\quad\qquad=\frac{(s-b)(s-c)}{bc}\ .\\[3mm] \frac r{2R} &=\frac{(s-b)(s-c)}{bc}\ ,\\ r\; bc &=2R\; (s-b)(s-c)\ ,\\ 2rs\;bc\; (s-a)&= 4R\;s(s-a)(s-b)(s-c)\ ,\\ 2S\;bc\; (s-a)&= 4R\;S^2\ ,\\ 2bc\; (s-a)&= 4RS\ ,\\ 2bc\; (s-a)&= abc\ ,\\ 2(s-a)&= a\ ,\\ 2s&=3a\ ,\\ a+b+c&=3a\ ,\\ b+c&=2a\ . \end{aligned} $$

$\square$