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$H$ is orthocenter of $\triangle ABC$. $J$ is a point on interior angle bisector of $\angle A$ that $HJ \perp AJ$.$J'$ is a point on exterior angle bisector of $\angle A$ that $HJ' \perp AJ'$.Prove the line that passes $J$ and $J'$ divides $BC$ to equal segments.

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Things I have done: $JJ'$ intersects $BC$ on $M$.I was able to show that $\angle HAJ = \angle JAO$. As $AJ'HJ$ is a rectangle we can conclude that $\angle HKJ = 2 \angle HAJ = \angle HAO$. So $AO\parallel KJ$. I stuck here. I know the theorem that distance between orhtocenter and vertex is twice as distance between circumcenter and side center(Proof of this can be found here),but i don't know how to apply it here.

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In a triangle, the orthocenter and the circumcenter are isogonal conjugates, hence $\widehat{HAJ}=\widehat{JAO}$ and the line through $K$ and $J$ is parallel to the $AO$-line ($JKA$ is an isosceles triangle). This implies that $AOMK$ is a parallelogram, since $AO\parallel JK$ and $AH\parallel MO\perp BC$.

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  • $\begingroup$ thanks for posting your solution.but we don't know $M$ is center of $BC$ .$JJ'$ just continued and intersected $BC$ in $M$.would you please tell how do deduced that $MO \perp BC$? $\endgroup$ Commented Sep 6, 2014 at 15:56
  • $\begingroup$ Take $M$ as the midpoint of $BC$. Then $AH\parallel MO\perp BC$. You know that $OM=AK$ and $KJ\parallel AO$, because $$\widehat{KJA}=\widehat{KAJ}=\widehat{JAO}.$$ $\endgroup$ Commented Sep 6, 2014 at 16:02

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