Using Radical Axis to prove Concurrence Let $BB',CC'$ be altitudes in $\triangle ABC$, and assume $AB\neq AC$. Let $M$ be the midpoint of $BC$, $H$ the orthocenter of $\triangle ABC$, and define $D$ as the intersection of lines $BC$ and $B'C'$. Show that $DH$ is perpendicular to $AM$
I am trying to apply the converse of the radical axis theorem to prove the desired prependicularity (since $AC'HQ$ cyclic implies $DQ'\perp AH$), but am having trouble on finding what circles to use.
The theorem states: Suppose that $ABCD$ and $CDEF$ are cyclic quadrilaterals and that line $AB,CD,EF$ are concurrent. Then $EFAB$ is also cyclic. More generally, if $\omega_1,\omega_2$ are two cricles with radical axis $l$, and $A,B$ are points on $\omega_1$, and $AB$ and $EF$ meet at a point on $l$, then $ABEF$ are concylic.
Anyone have any ideas?
 A: Not using the radical axis, and not even purely geometrical, but here is a relatively straightforward proof.
Assume WLOG that $A, B, C\,$ lie in a complex plane with the origin set to the circumcenter $O$ of $\triangle ABC$, and scaled such as $\lvert OA \rvert = \lvert OB \rvert = \lvert OC \rvert = 1$.
With the usual notation where $z$ is the complex number representing point $Z$, let $z = a + b + c$. Then  $\langle z - a, b - c \rangle = \langle b + c, b - c \rangle = \langle b, b \rangle + \langle b, c \rangle - \langle b, c \rangle - \langle c, c \rangle = {\lvert b \rvert}^2 - {\lvert c \rvert}^2 = 1 - 1 = 0$ thus $AZ \perp BC$ so $Z$ is on the altitude through $A$. By symmetry, Z must be on the other 2 altitudes of the triangle as well, thus $Z \equiv H$ the orthocenter of the triangle, and therefore:
$$
h = a + b + c
$$
(As a side note, the relation is well known in the context of the Euler line, though it is more commonly stated in vector notation $\overrightarrow{OH} = \overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}$, see for example paragraph "9. Vector Algebra II" at Existence of the Orthocenter.)
Let $A'$ be the foot of the altitude through $A$. Since $A'$ is on $BC$:
$$
a' = \lambda b + \mu c \;\;\;\; \text{for some real}\;\; \lambda, \mu \;\; \text{with} \;\; \lambda + \mu = 1
$$
Furthermore, since $A'H \perp BC$:
$$
\begin{align}
0 = \langle h - a', b - c \rangle & = \langle a + (1 - \lambda) b + (1 - \mu) c, b - c \rangle \\
  & = \langle a + \mu b + \lambda c, b - c \rangle \\
  & = \langle a, b \rangle - \langle a, c \rangle + \mu \langle b, b \rangle - \mu \langle b, c \rangle + \lambda \langle b, c \rangle - \lambda \langle c, c\rangle \\
  & = -\lambda + \mu + (\lambda - \mu) \langle b, c \rangle + \langle a, b \rangle - \langle a, c \rangle
\end{align}
$$
By construction, point $D$ is the harmonic conjugate of $A'$ with respect to $BC$ (easy to show using the Ceva/Menelaus theorems, for example). Since $AB \ne AC$ it follows that $\lambda \ne \mu$ and $D$ is a proper point (not at infinity) given by:
$$
d = \frac{\lambda b - \mu c}{\lambda - \mu}
$$
Finally, the midpoint $M$ of $BC$ is:
$$
m = \frac{b + c}{2}
$$
Now, it remains to be shown that $\langle h - d, m - a \rangle = 0$ in order to complete the proof.
$$
\begin{align}
\begin{split}
2 (\lambda - \mu) \langle h - d, m - a \rangle =& \langle (\lambda - \mu) a - \mu b + \lambda c, b + c - 2 a \rangle \\
  =& (\lambda - \mu) \langle a, b \rangle + (\lambda - \mu) \langle a, c \rangle - 2 (\lambda - \mu) {\lvert a \rvert}^2  \\
   &{} - \mu {\lvert b \rvert}^2 - \mu \langle b, c \rangle + 2 \mu \langle a, b \rangle \\
   &{} + \lambda \langle b, c \rangle + \lambda {\lvert c \rvert}^2 - 2 \lambda \langle a, c \rangle \\
  =& -\lambda + \mu + (\lambda + \mu) \langle a, b \rangle - (\lambda + \mu) \langle a, c \rangle + (\lambda - \mu) \langle b, c \rangle \\
  =& -\lambda + \mu + (\lambda - \mu) \langle b, c \rangle + \langle a, b \rangle - \langle a, c \rangle \\
\end{split}
\end{align}
$$
The last line matches the expression for $0 = \langle h - a', b - c \rangle$ calculated earlier. It follows that $\langle h - d, m - a \rangle = 0$, thus $DH \perp AM$. 

[ EDIT ]  The following elaborates on a couple of steps in the proof, as asked in a comment.


By construction, point $D$ is the harmonic conjugate of $A'$ with respect to $BC$

This is in fact the canonical construction for the projective harmonic conjugate point, which can be proven as follows. Since cevians $AA'$, $BB'$, $CC'$ are concurrent, it follows from Ceva's theorem that (using signed segments):
$$
\frac{BA'}{A'C} \cdot \frac{CB'}{B'A} \cdot \frac{AC'}{C'B} = 1
$$
Since points $B'$, $C'$, $D$ are collinear, it follows from Menelaus' theorem that (again using signed segments):
$$
\frac{BD}{DC} \cdot \frac{CB'}{B'A} \cdot \frac{AC'}{C'B} = -1
$$
Since the last 2 fractions are identical between the 2 products, it follows that:
$$
\frac{BD}{DC} = - \frac{BA'}{A'C}
$$
therefore $D$ is by definition the harmonic conjugate of $A'$ with respect to $BC$.

$D$ is a proper point (not at infinity) given by $d = \frac{\lambda b - \mu c}{\lambda - \mu}$

Let $Z$ be an arbitrary point on the line through $BC$ written as a linear combination $z = (\lambda b + \mu c) / (\lambda + \mu)$ with real $\lambda$, $\mu$. Then the signed segments ratio:
$$
\frac{BZ}{ZC} = \frac {z - b}{c - z} =  \frac{\lambda b + \mu c - \lambda b - \mu b}{\lambda c + \mu c - \lambda b - \mu c} = \frac{\mu (c - b)}{\lambda (c - b)} = \frac{\mu}{\lambda}
$$
Therefore to obtain the harmonic conjugate $Z'$ for which $\frac{BZ'}{Z'C} = - \frac{BZ}{ZC}$ it is enough to replace $\mu$ with $-\mu$ which gives $z' = \frac{\lambda b - \mu c}{\lambda - \mu}$.
