Proof for this property of triangles In a handout of mine a 'geometrical fact ' is stated :

Among such triangles ABC which have two fixed side lengths |BC|=a and |AC|=b<a, the triangle of largest angle ABC has BAC= 90°

Being a fact to solve physics problems it was not explained but I would like to know why this is true.
My attempt was to draw line segment BC and the locus of possible points of A (circle of radius b centered at C) and some how convince my self the tangent from B to this circle makes the greatest angle with BC
But I could not ,any way forward ?
 A: In $\triangle ABC$, length of $BC$ is greater that length of $AC$, hence the angle opposite to $BC$, $\angle BAC$ is greater than $\angle ABC$ which is the angle opposite to $AC$ $\; ( \therefore \;\angle BAC> \angle ABC\implies \angle ABC\;  \text{is acute} )$
By Sine Rule,
$$\frac{a}{\sin \angle BAC}=\frac{b}{\sin \angle ABC}\implies \sin \angle ABC=\frac{b}{a}\cdot \sin \angle BAC$$
Since $\frac{b}{a}$ is constant, $\angle ABC$ is maximum when $\sin \angle BAC=1\implies \angle BAC=90^{\circ}$
A: This is not meant to be a rigorous answer. Here is a nice simulation I made, inspired from the OP's graphical proof. One can see that $\angle B$ is maximized when $AB$ is tangent to the circle centred at $C$ and having radius $b$.

If anyone is interested, here is the Desmos link.
A: EDIT: Giving a direct calculus maximum calculation instead, with algebraic manipulation.
Sine Rule, usual trig notation
$$\frac{ \sin \beta}{\sin \gamma} =\frac{b} {\sqrt{a^2+b^2-2 ab \cos \gamma}} \tag 1$$
We have to maximize $\beta $ as a function of $ \gamma$.  As sine is monotonous in the interval under trig considerations we take square of sine with $x= \cos \gamma$ using quotient/chain rules for differentiation
$$ f(x)= \frac {1-x^2}{a^2+b^2-2ab x}= \frac{x}{ab} \tag 2 $$
Cross multiply, simplify with
$$q=\frac{a^2+b^2}{2ab} \tag3 $$
to solve quadratic equation $$x^2- 2q x +1=0\to x= a/b,b/a \tag 4 $$
from which the latter is discarded by given condition $a>b$
$$ \cos \gamma = b/a ;\quad c= {a^2+b^2-2ab \cdot b/a} =\sqrt{a^2-b^2} \tag 5 $$
They make a Pythagorean triplet $a^2= b^2+c^2$ making $\angle BAC  =  90 ^\circ. $
