Formula for distance between Incenter and Orthocenter. On pg. 205 of Johnson's Advanced Euclidean Geometry, he gives three formulas, preparatory to a proof of Feuerbach's theorem.
$$OI^2=R^2-2R\rho$$
$$IH^2=2\rho^2-2Rr$$
$$OH^2=R^2-4Rr$$
where $O$ is the circumcenter, $I$ the incenter, $H$ the orthocenter; $R$ the radius of the circumcircle, $\rho$, that of the inscribed circle, and $r$, the in-radius of the pedal triangle, the triangle formed by the feet of the altitudes.
The first equation is the theorem of Euler. The third follows from the first, by applying it to the pedal triangle, whose incenter is $H$, and circumcenter is $N$, the midpoint of $HO$.
Is there a correspondingly simple proof of the second formula? A non-trigonometric solution is preferred.
 A: Here is a try to get the formula quickly.

Disclaimer: I will use more or less standard notations in a triangle $\Delta ABC$. The side lengths are $a,b,c$, the incenter is $I$, the inradius is $r$ (not $\rho$), the circumradius is $R$, the area is $S$. Let $AA'$, $B'$, $CC'$ be the heights intersecting in $H$.  The inradius of the $H$-pedal triangle $\Delta A'B'C'$ will be denoted by $r_H$.

The barycentric coordinates of $I$ are $I=[a:b:c]$, explicitly we have (identifying with affixes)
$$
(a+b+c)I=aA+bB+cC\ .
$$
Then considering vectors:
$$
\begin{aligned}
a(a+b+c)\overrightarrow{IA} &= 
a^2\overrightarrow{AA} +
ab\overrightarrow{BA} +
ac\overrightarrow{CA} 
\ ,\\
b(a+b+c)\overrightarrow{IB} &= 
ba\overrightarrow{AB} +
b^2\overrightarrow{BB} +
bc\overrightarrow{CB} 
\ ,\\
c(a+b+c)\overrightarrow{IC} &= 
ca\overrightarrow{AC} +
cb\overrightarrow{BC} +
c^2\overrightarrow{CC} 
\ ,
\end{aligned}
$$
so after adding:
$$
\tag{$1$}
a\overrightarrow{IA}+
b\overrightarrow{IB}+
c\overrightarrow{IC} =0\ .
$$

Write now $\overrightarrow{HA}=\overrightarrow{HI}+\overrightarrow{IA}$, take norms, and multiply with $a$. Write the similar equalities for $B$, $C$ instead of $A$ to get:
$$
\begin{aligned}
aHA^2 &=aHI^2 + aIA^2 + 2a\overrightarrow{HI}\cdot \overrightarrow{IA}\ ,\\
bHB^2 &=bHI^2 + bIB^2 + 2b\overrightarrow{HI}\cdot \overrightarrow{IB}\ ,\\
cHC^2 &=cHI^2 + cIC^2 + 2c\overrightarrow{HI}\cdot \overrightarrow{IC}\ .
\end{aligned}
$$
Adding, we get rid of the scalar products by $(1)$, and obtain:
$$
\tag{$2$}
aHA^2+bHB^2+cHC^2 = (a+b+c)HI^2 
+ \underbrace{aIA^2+bIB^2+cIC^2}_{\text{some constant }K^3}\ .
$$

Note: We did not use any special property of $H$, so $(2)$ is valid for any point $X$ instead of $H$, explicitly
$$
\tag{$2_X$}
aXA^2+bXB^2+cXC^2 = (a+b+c)XI^2 + K^3\ .
$$
(Key-word: Leibnitz formula.)
(For $I$ we have used only $I=[a:b:c]$. So we can also replace $I$ by a general point $P=[u:v:w]$, using the corresponding barycentric  coordinates $u,v,w$ instead.)

Now compute the relation $a(2_A)+b(2_B)+c(2_C)$ divided by $(a+b+c)$, weights being used to make appear on the R.H.S. the term $\sum a AI^2=K^3$.
So after the sum we have on the R.H.S. $K^3+K^3$.
On the L.H.S. we get $\frac 1{a+b+c}\sum(abc^2+acb^2)=2abc$.
This gives the value for $K^3$,
$$
K^3 = abc = 4R\;S=4R\; rp=2Rr\;(a+b+c)\ .
$$
Here $p=\frac 12(a+b+c)$ is the half perimeter. Inserting this into $(2)$ we get
$$
\tag{$3$}
\frac1{a+b+c}(aHA^2+bHB^2+cHC^2) =  HI^2 
+ \underbrace{2Rr}_{abc/(a+b+c)}\ .
$$

We want now to show $(!)$ the following similar relation:
$$
\tag{$4$}
2r^2 \overset != HI^2 
+ 2Rr_H\ .
$$
Showing it is equivalent to:
$$
\tag{$4'$}
\frac1{2p}(aHA^2+bHB^2+cHC^2) - \frac{abc}{2p}
\overset !=  
2r^2 - 2Rr_H\ .
$$

It is time to write down some trigonometric formulas.
$$
\begin{aligned}
BC &= a\ ,\\
B'C' &= BC\cos A = a\cos A &&\text{ from }\Delta ABC\sim \Delta AC'B'\ ,\\
AH &= AB'\sin\widehat{AHB'}=c\cos A\sin C= 2R\cos A\ ,\\
HA'&=HB\cos\widehat{AHB'} =2R\cos B\cos C\ ,\\
r_H &=HA'\cdot \frac{B'C'}{BC}=2R\cos A\cos B\cos C
&&\text{ from }\Delta HB'C'\sim \Delta HCB\ ,\text{ so}\\
2Rr_H &= HA\cdot HA'=HB\cdot HB'=HC\cdot HC'\ .
\end{aligned}
$$

I understand that trigonometry should be avoided, but then computations
are hard to write. Above, there are also some alternative relations showing how to avoid - if really needed - trigonometric formulas. At any rate, no trigonometric relations will be used below. (E.g. like writing $\cos^2A$ in terms of $\cos 2A$ or so.)
We want to show $(4')$ and are in position to show this as an algebraic formula in the variables $a,b,c$. For this, we use for $S^2$ Heron, and $R$ is eliminated via $abc=4RS$.
The version of $(4')$ suited for this polynomial check is:
$$
\tag{$4''$}
\frac1{2p}4R^2(a\cos^2A + b\cos^2B + c\cos^2 C) - \frac{abc}{2p}
\overset !=  
\frac{2S^2}{p^2} - 4R^2\cos A\cos B\cos C\ .
$$
Equivalently:
$$
\tag{$4'''$}
\frac1{2p}4R^2(a\cos^2A + b\cos^2B + c\cos^2 C) 
+
\frac1{2p}4R^2(2p\; \cos A\cos B\cos C)
\overset !=  
\frac{abc+4(p-a)(p-b)(p-c)}{2p} \ .
$$
And we multiply with $2p$.

I have to finish, so a computer check is done for the above.
var('a,b,c')
p = (a + b + c)/2
SS = p*(p - a)*(p - b)*(p - c)
RR = (a*b*c/4)^2 / SS
def f(a, b, c):    return (b^2 + c^2 - a^2)/2/b/c
cosA, cosB, cosC = f(a,b,c), f(b,c,a), f(c,a,b)

RHS = a*b*c + 4*(p - a)*(p - b)*(p - c)
LHS = 4*RR*(a*cosA^2 + b*cosB^2 + c*cosC^2 
            + 2*p*cosA*cosB*cosC)

print(bool(LHS == RHS))

And the computer checks the relation. Explicitly:
sage: print( bool(LHS == RHS) )
True
sage: 2*RHS.expand()
-a^3 + a^2*b + a*b^2 - b^3 + a^2*c + b^2*c + a*c^2 + b*c^2 - c^3
sage: 2*LHS.factor().expand()
-a^3 + a^2*b + a*b^2 - b^3 + a^2*c + b^2*c + a*c^2 + b*c^2 - c^3

(I will search for a better finish
based on geometric relations.)
