# Some geometry - I have found a paralelogram

The $ABC$ acute triangle's orthic triangle is the $DEF$ triangle, $D\in AB, E\in AC, F\in BC$. Let $F'$ and $F"$ be points on $DE$, such that $EF'=EF, DF"=DF$ and neither $F'$ nor $F"$ lie inside the circumcircle of triangle $ABC$. Let the center of the circumcircle $AF'F"$ be point $O$. Furthermore let $O'$ be the circumcenter of trianle $ABC$, $e$ be the line tangent to the circumcircle of $\triangle ABC$ at point $A$ and $X\neq A$ be the intersection of $AB$ line and the circumcircle of $AF'F"$. Prove that lines $e, DE, O'B, OX$ define a paralelogram.

Well, I proved that $F'$ and $F"$ is the reflection of $F$ to $AB, AC$. Note that the $AF'F"$ is an isosceles triangle with $\angle F'AF" = 2\times \angle BAC.$ Maybe I should join the perpendicular bisector of $F'F"$? How can I prove it?

• It's really hard to see what you're asking, because your figure has no points labeled. – Steve Kass Jan 19 '18 at 20:33
• I changed the figure. Is it OK now? – Pet123 Jan 20 '18 at 8:59

Firstly, you have shown that $\triangle AF'F''$ is isosceles. You may want to note that $\angle AF''F' (\angle AF''D)=\angle AFD=\angle EBA$

You may also note that $\angle EBC=\angle FAE=\angle EAF'$

Combining these two, we have $\angle AF''F'+\angle F'AE=\angle EBC+\angle EBA$.

Let $O'B$ intersect $e$ at $T$.

$e$ is tangent to circle $ABC$ implies that $\angle EAT=\angle CBA$ which also means that $\angle TAF'+\angle F'AE=\angle EBA+\angle EBC$.

Now we see that $\angle TAF'=\angle AF''F'$. This means that $e$ is tangent to circle $AF''F'$ too. And since $\triangle AF''F'$ is isosceles, that means that $e$ is parallel to $F'F''$. Half the problem is done.

Now we just have to show $O'B\parallel OX$. But by above, circle $ABC$ internally tangent to circle $AF'F''$ at $A$, so $\frac{AB}{AX}=\frac{r}{R}=\frac{AO'}{AO}$. Also, $A,O',O$ are collinear.(These are very well-known properties of internally tangent circles). And the result follows.