Prove that $r^3\rho=2R \rho_1\rho_2 \rho_3$ Let $ABC $ be an acute-angled triangle.
Let $D$, $E$, $F$ be the feet of the perpendiculars from $A $, $B$, $C $ on the opposite sides $BC $, $CA $, $AB $. Let $\rho,\rho_1,\rho_2 ,\rho_3$ be the radii of the circles inscribed in the triangles $DEF$, $AEF$, $BFD$, $CDE$.
Prove that $r^3\rho=2R \rho_1\rho_2 \rho_3$.
Here, $r $ is the radius of the incircle of triangle $ABC$, and $R$ is the circumradius of $ ABC$.
 A: Recall that the $\triangle ABC$ is acute. 
Few (easy) observations:

*

*$H$ is the center of the incircle of $\triangle DEF$ -- $H,E,B,F$ lie on a circle, so $\angle HDF=\angle HBF=90^\circ - A$. Similarly, $\angle HDE=\angle HCE=90^\circ-A$.

*$\triangle BDF$ is similar to $\triangle BAC$ -- $BD:AB=BF:BC=\cos B : 1$.

*$BH$ is the diameter of the circumcircle of $\triangle BDF$ -- $\angle HFB=\angle HDB = 90^\circ$.

Now $\rho_2:r = BD:AB$, so $\rho_2=r\cos B$. Thus we have
$$\rho_1\rho_2\rho_3 = r^3\cos A\cos B\cos C.$$
So we need to show that
$$\rho = 2R\cos A\cos B\cos C.\tag{1}$$
Let $G $ be the foot of the perpendicular from $H $ on $DE $.
Since $$\rho=HG=HD\sin(\angle HDE)=HD\sin(90^\circ-A)=HD\cos A,$$
and
$$HD=BH\sin(\angle HBD)=BH\sin(90^\circ -C) = BH\cos C,$$
Equality (1) becomes
$$BH\cos A\cos C=2R\cos A\cos B\cos C$$
which is equivalent to
$$BH=2R\cos B.\tag{2}$$
But from Observation 2 and 3 we have
$$BH:2R = BD:AB =\cos B:1.$$
Thus Equality (2) holds. QED
A: Since the geometical calculus is rather bulky, the sheet below includes only the main equations :


In order to save space and time, the known formulas of radii are not repeated here. References :
$[1]$ : http://mathworld.wolfram.com/Circumradius.html   Equation (1)
$[2]$ : http://mathworld.wolfram.com/Incircle.html   Equation (7)
Thanks very much to have pointed out some typos in the attached sheet. They are corrected.
