$AB_1$, $AB_2$, $AB_3$ are the altitude, angle bisector, median from vertex $A$ of $\triangle ABC$; arrange lengths $BB_i$ in ascending order 
Consider an acute angled triangle $\triangle ABC$ such that $AB\lt AC$.

If from $A$ altitude $AB_1$ is drawn, internal angle bisector $AB_2$ is drawn, and median $AB_3$ is drawn.
Arrange the lengths $BB_1$, $BB_2$ and $BB_3$ in ascending order.

My try: I started with an Isosceles Triangle $\triangle ABD$ with $AB=AD$.

Now, for $\triangle ABD$, $AB_1$ is altitude, angle bisector, and median.
In figure $2$
Let $\angle BAB_1=\theta=\angle B_1AD$
let $\angle DAC=2 \beta$
So $\angle BAC=2(\theta+\beta)$
If we construct $AB_2$ asinternal angle bisector of $\angle BAC$, Then each half angle is :
$$\angle BAB_2=B_2AC=\theta+\beta \gt \theta$$
$\implies$
$$\angle BAB_2 \gt \angle BAB_1$$
hence the point $B_2$ should be to right side of the point $B_1$
Hence $$BB_1 \lt BB_2$$
But can I have a clue to compare $BB_2$ and $BB_3$?

$BB_2:B_2C = AB:AC < 1$ so $BB_2 < BC/2 = BB_3$.
