0
$\begingroup$

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? enter image description here

$\endgroup$
  • 1
    $\begingroup$ It's really hard to see what you're asking, because your figure has no points labeled. $\endgroup$ – Steve Kass Jan 19 '18 at 20:33
  • $\begingroup$ I changed the figure. Is it OK now? $\endgroup$ – Pet123 Jan 20 '18 at 8:59
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.