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.
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 $S'=S$.)
$\square$
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.
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 $a,b,c$.
$$
\begin{aligned}
AI&\perp OI\ ,\\
AI^2 &=AO^2-OI^2\ ,\\
AI^2&= R^2-(R^2-2Rr) \ ,\\
AI^2 &= 2Rr\ ,\\
\frac r{2R} &=\frac{r^2}{AI^2}\ .\\[3mm]
&\quad\text{Separated computation:}\\
&\quad\frac{r^2}{AI^2}=\sin ^2\frac A2=1-\cos ^2\frac A2
=\frac {1-\cos A}2
=\frac 12\left(1-\frac{b^2+c^2-a^2}{2bc}\right)\\
&\quad\qquad=\frac 1{4bc}(a^2-(b-c)^2)=\frac 1{4bc}(a-b+c)(a+b-c)
\\
&\quad\qquad=\frac{(s-b)(s-c)}{bc}\ .\\[3mm]
\frac r{2R} &=\frac{(s-b)(s-c)}{bc}\ ,\\
r\; bc &=2R\; (s-b)(s-c)\ ,\\
2rs\;bc\; (s-a)&= 4R\;s(s-a)(s-b)(s-c)\ ,\\
2S\;bc\; (s-a)&= 4R\;S^2\ ,\\
2bc\; (s-a)&= 4RS\ ,\\
2bc\; (s-a)&= abc\ ,\\
2(s-a)&= a\ ,\\
2s&=3a\ ,\\
a+b+c&=3a\ ,\\
b+c&=2a\ .
\end{aligned}
$$
$\square$
Best Answer
From inscribed angle theorem we have the following observations:
With respect to the circle $O_1$:
$$\angle BCA=\angle BAD. $$ With respect to the circle $O_2$: $$ \angle BDA=\angle BAC. $$
It follows $\triangle ABD\sim \triangle CBA$. This implies: $$ \frac{CB}{AB}=\frac{r_1}{r_2}>1\implies CB>AB,\quad \frac{CB}{AB}=\frac{AB}{BD}\implies AB=\sqrt{CB\cdot BD}<\frac{CB+BD}2, $$ where the last is AM-GM inequality.