# 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 $OA'= R \cos A$?

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..What is $A'$? –  Aryabhata Apr 17 '12 at 20:48
I'm guessing that $A^{\prime}$ is the midpoint of $BC$ (and the foot of the perpendicular from $O$ to $BC$). In this case, $\angle BOA^{\prime} = (1/2)\angle BOC = \angle A$; the first equality is clear, and the second equality is the Inscribed Angle Theorem (en.wikipedia.org/wiki/Inscribed_angle#Theorem ) with respect to the circumcircle. Then, $OA^{\prime}$ is the leg adjacent to $\angle BOA^{\prime}$ in a right triangle with hypotenuse $R$, so ... –  Blue Apr 17 '12 at 21:07

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.

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