Let $\triangle ABC$ be an acute-angled triangle. Let $H$ be the foot of the perpendicular from A to BC. Let $K$ be the foot of the the perpendicular of $H$ to $AB$, let $L$ be the foot of the perpendicular from $H$ to $AC$. Let $AH$ intersect the circumcircle of $ \triangle ABC$ in $T$, and let the line through $K$ and $L$ intersect the circumcircle of $\triangle ABC$ in $P$ and $Q$. Prove that $H$ is the incenter of $\triangle PQT$.

enter image description here

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    $\begingroup$ You seem to be using same names for both points and lines. Could you clarify what is what in this problem? $\endgroup$
    – Wojowu
    Jan 5, 2015 at 22:26
  • $\begingroup$ Ah yes, I'll edit it $\endgroup$ Jan 5, 2015 at 22:31
  • $\begingroup$ You must assume that $\triangle ABC$'s non-acute angle, if any, is at $A$. If $B$ or $C$ is obtuse, then $H$ is actually an excenter of $\triangle PQT$. $\endgroup$
    – Blue
    Jan 6, 2015 at 10:48
  • $\begingroup$ @Blue yes, can you solve it? $\endgroup$ Jan 6, 2015 at 10:55
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    $\begingroup$ @wowlolbrommer: I haven't solved it yet. Where did you get this problem? From a textbook, or perhaps from a contest? Knowing the difficulty level could be helpful in deciding how to approach the problem. $\endgroup$
    – Blue
    Jan 6, 2015 at 10:57

1 Answer 1


In any triangle the orthocenter and the circumcenter are isogonal conjugates. Since $AKHL$ is a cyclic quadrilateral, this implies that the perpendicular to $PQ$ through $A$ passes through the circumcenter of $ABC$, $O$. This implies $\widehat{POA}=\widehat{AOQ}$, so $\widehat{PTA}=\widehat{ATQ}$, and $AP=AQ$. Let $A'$ be the symmetric of $H$ with respect to $A$. If we manage to prove $AP=AH$, then the perpendicular $l_P$ to $HP$ through $P$ and the perpendicular $l_Q$ to $HQ$ through $Q$ meet on $A'$, so $PTQ$ is the orthic triangle of the triangle $\Delta$ delimited by $l_P,l_Q$ and the perpendicular $l_T$ to $HT$ through $T$. This gives that $H$ is the orthocenter of $\Delta$, hence the incenter of $PTQ$.

enter image description here

The best way I found so far to prove $AP=AH$ is to use some trigonometry, but I think there are more clever ways.

Edit: Thanks to wowlolbrommer, here it is a clever way: $AP=AQ$ implies $\widehat{AQL}=\widehat{ACQ}$, hence $AQ^2=AL\cdot AC$. By Euclid's first theorem, $AL\cdot AC = AH^2$, hence $AP=AQ=AH$ and we're done.

  • $\begingroup$ if my figure is any guide, i don't that $AP = AH.$ $\endgroup$
    – abel
    Jan 6, 2015 at 16:41
  • $\begingroup$ @abel: I do not understand you. It is true that $AP=AH$, the only issue is that I have not found a nice proof of it, yet. $\endgroup$ Jan 6, 2015 at 16:43
  • $\begingroup$ AH/AC=AL/AH and AQ/AL=AC/AQ so AH=AQ=AP $\endgroup$ Jan 7, 2015 at 0:10
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    $\begingroup$ Assuming the result is true, then it's definitely also true that $AP=AH$. This is, by interesting coincidence, the upshot of the Lemma in my answer to the recent Japanese Theorem question. $\endgroup$
    – Blue
    Jan 7, 2015 at 0:34
  • $\begingroup$ nice. AHC and ALH are similar for the first equality, AQL and ACQ re similar for the 2nd: QAL=CAQ, LQA=PQA=APQ=QCA $\endgroup$ Jan 7, 2015 at 6:32

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