Euler line of triangle $ABC$ is parallel to side $BC$ $\implies$ $\tan B \tan C = 3$ I'm having some trouble on this exercise from Geometry Revisited:

On triangle $ABC$, the Euler line is parallel to $BC$. Prove that $\tan B \tan C = 3$.

Here is the solution given:

(in this context, $D$ is the base of the altitude from $A$, $O$ is the circumcenter, $A'$ is the midpoint of $BC$, and $R$ is the radius of the circumcircle)
I understand everything up to the last line. However, I do not understand how the line $OA'$ is related to the circumradius.
Why is $OA'= R \cos A$?
 A: Okay, I figured out what was going on after Day Late Don posted his comment: since he didn't write a complete answer I'll answer this myself.
I probably should have included this diagram when I asked the question:

Since $O$ is the circumcenter, we have $\angle BOA' = \frac{1}{2} \angle BOC = \angle A$, so $OA' = R \cos A$. Since $\cos A = \cos (180-B-C)$, this side is equal to $R(\sin B \sin C - \cos B \cos C)$ by some trig identities.
Now we know that $OA' = \frac{1}{3} AD$, so we have the following equation:
$$R(\sin B \sin C - \cos B \cos C) = \frac{1}{3} (2R \sin B \sin C)$$
Simplifying this yields the desired tangent equality.
A: Coincidentally I was recently working the same problem too; Here's another version of this solution, with the tangent multiple equalling 3 and the Euler line OH shown. 

Here is another diagram about which I had questions. I dragged vertex B from its position in the diagram above to form an obtuse triangle. Now the product of the tangents is -3, and the Euler line is no longer parallel (or orthogonal). [The measured angle OLB gives the angle; The label positions are off in the diagram but L is the point to the right, that looks like J because of the label placement]. It seems more than just random that the tangent product is -3, but maybe not. I haven't tried to figure out if there is another tangent product statement that could be made here...

