Consider an acute angled triangle $\triangle ABC$ such that $AB\lt AC$.

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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 $AB=AD$.

enter image description here

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 :

$$\angle BAB_2=B_2AC=\theta+\beta \gt \theta$$


$$\angle BAB_2 \gt \angle BAB_1$$

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. B123

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 $BB_3.$


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