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Let $\triangle ABC$ be a acute-angled triangle so that $AB>AC$ and $\angle BAC = 60^\circ$. Let $O$ be the circumcenter and $H$ the orthocenter. Let $OH$ intersect $AB$ and $AC$ in $P$ and $Q$ respectively. Show that $AQ=AP$. Or alternatively, show that $\angle AOP$ and/or $\angle AHQ$ equal $\angle ACB+90-\angle BAC$

I know $BAO=CAH=90-\angle ACB$, but that's all I've got. I need one of two above. An elementary solution is preferred.

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  • $\begingroup$ As $AH= 2R\cos A$, in this case $\cos A=\frac12$, so $AH=AO$. The other angles at the vertex are easily proven equal, so $\triangle APO \cong \triangle AQH$. Simple :D $\endgroup$ – Sawarnik Apr 6 '15 at 19:28
  • $\begingroup$ Did you know that thing, $AH= 2R\cos A$? Otherwise we would have to think of a geometric proof that $AH=AO$. :D $\endgroup$ – Sawarnik Apr 6 '15 at 19:30
  • $\begingroup$ No where does that come from? $\endgroup$ – steedsnisps Apr 6 '15 at 19:31
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    $\begingroup$ Hmm, let CH cut AB at D. Then, $AD= b\cos A$. So, $AH= \frac{AD}{\sin B} = 2R\cos A$, by sine rule. Do you follow? :D $\endgroup$ – Sawarnik Apr 6 '15 at 19:36
  • $\begingroup$ Yes. It's not exactly elementary, though. $\endgroup$ – steedsnisps Apr 6 '15 at 19:58
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enter image description here

Construction: CH is produced to cut the circumference at X. BX, when joined, cuts QP produced at M.

The pink angles are all $30^0$.

OM is the perpendicular bisector of BX. This makes the angles marked red are all equal to $30^0$.

The above is sufficient to say $\angle AQH = \angle APQ = 60^0$

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    $\begingroup$ Is $OM$ the perpendicular bisector of $BX$ because $|OX|=|OB|$ and $|HX|=|HB$| so $H$ and $O$ lie on the perpendicular bisector of $BX$? $\endgroup$ – steedsnisps Apr 6 '15 at 20:53
  • $\begingroup$ @wowlolbrommer By “angles in the same segment”, $\angle X = \angle BAC = 60^0$. This makes ⊿HXB equilateral. Therefore, |HX|=|HB|. Together with your finding, also HO being the common side, we can say that ⊿HXO is congruent to ⊿HBO. This further means $\angle XHM = \angle BHM$. A step further means |XM| = |MB| and $\angle XMH = \angle BMH = 90^0$. Sorry for making a big jump. $\endgroup$ – Mick Apr 7 '15 at 8:51
  • $\begingroup$ ok thanks for your solution! $\endgroup$ – steedsnisps Apr 7 '15 at 12:02
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The length of $AH$ in any $\triangle ABC$ is given by $2R\cos A$, and can be proven rather easily using the sine rule. Thus, in this case $\cos A=\frac12$, thus $AH =OH$. Hence, $\angle POA=\angle QHA$.

Of course, $\angle OAP = \angle OAB = 90^{\circ}-\angle C=\angle HAC = \angle HAQ$. Thus, by virtue of ASA criterion, $\triangle APO \cong \triangle AQH$.

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  • $\begingroup$ But you're still using trig $\endgroup$ – steedsnisps Apr 7 '15 at 14:09
  • $\begingroup$ Ah .. let me think of a solution using similarity. $\endgroup$ – Sawarnik Apr 7 '15 at 16:50
  • $\begingroup$ It's alright otherwise :) $\endgroup$ – steedsnisps Apr 7 '15 at 18:31

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