Segment theorem joining the incenter to the barycenter

Question

In a right triangle, the segment whose ends are the incenter and the barycenter is parallel to one of the legs. Calculate the measure of one of its acute angles.(Answer:$37^o$)

There is a theorem that says the only right triangle where it is true that the segment that joins the incenter and centroid is parallel to a leg, is the one at 37° and 53°.

Can someone prove this theorem? I try but i didin't.

$GI \parallel BC \implies IH \perp AB\\ \angle A + \angle C = 90^o\\ \angle IAC +\angle ICA = \frac{\angle A}{2}+\frac{\angle C}{2}=45^o \implies \angle AIC = 135^o$

Answer 1

Here is a solution using analytic geometry, with the natural coordinates shown in the figure below (we assume $a \ge b$).

Let $G$ be the centroid of right triangle $AOB$, $I$ its incenter and $r$ its inradius.

There exists a formula for $r$ in our case (right triangle) (see proof by David Quinn here) :

$$r=\frac12(a+b-\sqrt{a^2+b^2})\tag{1}$$

Let us express the desired parallelism by writing that the ordinates of point $I$ and point $G$ are the same :

$$r=\frac{b}{3}$$

giving equation :

$$3(a+b-\sqrt{a^2+b^2})=2b\tag{2}$$

Isolating in (2) the square root and taking the square of both sides gives :

$$4ab=3a^2 \ \iff \ \frac{b}{a}=\frac{3}{4}=\tan(\angle OAB)$$

which means that

$$\angle OAB = \tan^{-1} \frac{3}{4} \approx 37°$$

Answer 2

Choose $K$ on $AB$ such that $GK\parallel AC$. As $GF/BF=1/3$ then we have $AK=1/3 AB$.

But triangle $HGK$ is similar to $ABC$: from $GK=1/3 AC$ it follows $HK=1/3 AB$.

Consequently: $BH=1/3 AB$, $AT=2/3 AB$, $CT=BC-1/3 AB$ and $AC=1/3 AB+BC$. From Pythagoras theorem we then have: $$ AB^2+BC^2=(1/3 AB+BC)^2, \quad\text{whence:}\quad AB/BC=3/4. $$