# $AB_1$, $AB_2$, $AB_3$ 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 $$BB_2 < BC/2 = BB_3$$.
We can perceive $$ABC$$ as a half of a parallelogram $$ABDC$$ with diagonals $$AC, BD.$$
Consider a rhombus $$ABD'C'$$ where $$C'\in BC$$ and $$AD',BC'$$ are diagonals. Denote $$B_1', B_2', B_3'$$ the points considered in the question and related to this rhombus.
Diagonals in a rhombus are perpendicular, are angle bisectors of the rhombus, and meet in their common midpoint (as it is for arbitrary parallelogram). Hence the points $$B_1', B_2', B_3'$$ coincide.
Move $$C'$$ along $$BC$$ towards $$C$$ keeping a parallelogram with the sides $$AB\;\text{and}\; AC.$$ Clearly, $$B_1'$$ will not move while $$B_2'$$ and $$B_3'$$ do.
Return to the notation $$C,B_1,B_2,B_3.$$
The angle $$\angle AB_3C$$ becomes obtuse, while $$\angle BB_3C$$ is acute. Consequently, $$\angle BAB_3 < \angle B_3AC.$$ Since $$AB_2$$ is the angle bisector, $$B_2$$ lies on $$BB_3.$$