Which is the smallest acute angle of a right triangle in which the segment that joins the incenter with the barycenter is parallel to one side.? For reference:Calculate the smallest acute angle of a right triangle in which
the segment that joins the incenter with the barycenter is parallel to one side.
My progress:
I solved this question based on the properties of incenter and barycenter in the right triangle
There are 2 properties that say:

*

*The only right triangle in which the segment that joins the incenter and the barycenter and is parallel to a side, is the one with 37° and 53°.

*When the line that joins the barycenter to the incenter of any triangle is parallel to one of the sides, we have that the sides of this triangle are in P.A (arithmetic progression).

but could anyone demonstrate this property?
 A: Let $a, b, c$ be the sides of $\triangle ABC$.
WOLOG, we will assume $a \le b \le c$ and $C$ is the right angle.
Let $(u,v,w)$ be barycentric coordinates associated with $\triangle ABC$.
In barycentric coordinates, lines parallel to the sides have the form:

*

*parallel to $AB \longrightarrow w = $ constant.

*parallel to $BC \longrightarrow u = $ constant.

*parallel to $CA \longrightarrow v = $ constant.

Let $\ell = a + b + c$ be the perimeter. The barycentric coordinates for incenter $I$ is $(\frac{a}{\ell}, \frac{b}{\ell}, \frac{c}{\ell})$ and barycenter/centroid $G$ is $(\frac13,\frac13,\frac13)$. In order for $I$ and $G$ to lie on a line parallel to a side, one of $\frac{a}{\ell}, \frac{b}{\ell}, \frac{c}{\ell}$ need to equal to $\frac13$.
It is clear this cannot be $\frac{a}{\ell}$ nor $\frac{c}{\ell}$. Otherwise, the inequality $a \le b \le c$ will force $a = b = c$, contradict with the fact $\triangle ABC$ is not an equilateral triangle.
This implies $b = \frac{\ell}{3}$.
Since $a^2 + b^2 = c^2$, this leads to a unique solution of $a,b,c$ (up to scaling):
$$a^2 + \left(\frac{\ell}{3}\right)^2 = \left(\frac{2\ell}{3} - a\right)^2\quad\implies\quad
(a,b,c) = \left(\frac{\ell}{4},\frac{\ell}{3},\frac{5\ell}{12}\right)$$
The sides are in the ratio $a : b : c = 3 : 4 : 5 $ (an arithmetic progression) and
$$\begin{align}
\angle A &= \arcsin\frac{a}{c} = \arcsin\frac{3}{5} \sim 36.87^\circ\\
\angle B &= \arcsin\frac{b}{c} = \arcsin\frac{4}{5} \sim 53.13^\circ
\end{align}$$
A: 
Here is a diagram showing $GI$ parallel to one of the perpendicular sides. Given it is a right triangle, using standard notations for side lengths,
Inradius $r = \frac12 (b+c-a)$
But as $GI \parallel AB$, $r = AH = \frac{b}{3}$
Equating both,
$3 (b+c-a) = 2b \implies b + 3c = 3a$
Squaring both sides, $b^2 + 9c^2 + 6 bc = 9b^2+9c^2$
That leads to $4b = 3c ~ $ or $ ~b:c:a = 3:4:5$
You can also show that $GI$ cannot be parallel to hypotenuse.
