A relationship involving mutual distances between the feet of the median, the altitude and the angle bissector

Question

Consider a triangle $ABC$ with sidelengths $a:=BC, b:=AC,c:=AB$.

Let $M,D,H$ be resp. the feet of the median, the angle bisector and the altitude issued from $A$.

I have found this formula (see my proof below) :

$$4MD.MH=(b-c)^2 \tag{1}$$

My questions :

1. Has this formula been referenced somewhere ?

2. Are there alternative proofs (hopefuly simpler) ?

Proof :

This formula being homogeneous, we can assume WLOG that $a=2$. Let us take for the $x$-axis line $AB$, $M$ as the origin. In this way $A=(-1,0)$ and $B=(1,0)$.

We are going to establish formula (1) under the form :

$$16x_D^2.x_H^2=(b-c)^4 \tag{2}$$

where $x_D,x_H$ resp. are the abscissas of $D$ and $H$.

First of all, it is known that the foot $D$ of the angle bissector divides the side on which it is situated in the ratio of the other sides, which means here that the abscissa of $D$ is such that :

$$1-x_D=2\frac{b}{b+c} \iff x_D=\frac{c-b}{c+b}\tag{3}$$

Therefore, formula (2) can be written :

$$16x_H^2=(b^2-c^2)^2 \tag{4}$$

Let us recall two formulas :

• Heron's formula expressing the square of the area $\mathfrak{A}^2$ of triangle $ABC$ under the following form (we have taken into account the fact that $a=2$) :

$$\mathfrak{A}^2=\frac{1}{16}((b+c)^2-4)(4-(b-c)^2)\tag{5}$$

Besides, if $h:=AH$, the area of the triangle is as well given by :

$$\mathfrak{A}=\frac12 2 h = h\tag{6}$$

• the formula giving the square of the length of median $m:=AM$ :

$$m^2=\frac14\left(2(b^2+c^2)-\underbrace{a^2}_{4}\right)\tag{7}$$

Pythagoras' theorem in right triangle $AHM$ gives :

$$x_H^2=m^2-h^2=\frac14\left(2(b^2+c^2)-4\right)-\mathfrak{A}^2\tag{8}$$

Putting all this together, (4) will be established if, for all $a,b$ :

$$4 \left(2(b^2+c^2)-4\right)-((b+c)^2-4)(4-(b-c)^2)=(b^2-c^2)^2\tag{9}$$

which is indeed an identity.

Remark : The RHS of formula (1) is in relationship with a recent question here.

Answer 1

Are there alternative proofs (hopefuly simpler) ?

$x_H$ can be found more easily.

Pythagoras' theorem in right triangles $\triangle{AHB}$ and $\triangle{AHC}$ gives $$\bigg(AH^2=\bigg)\ c^2-(x_H+1)^2=b^2-(1-x_H)^2\tag1$$ (note that this holds even when $|x_H|\geqslant 1$)

Solving $(1)$ gives $x_H=\dfrac{c^2-b^2}{4}$.