Let $\triangle ABC$ be an acute triangle with incenter $I$ and circumcenter $O$. The incircle touches sides $BC,CA,$ and $AB$ at $D,E,$ and $F$ respectively, and $A'$ is the reflection of $A$ over $O$. The circumcircles of $ABC$ and $A'EF$ meet at $G$, and the circumcircles of $AMG$ and $A'EF$ meet at a point $H\neq G$, where $M$ is the midpoint of $EF$. Prove that if $GH$ and $EF$ meet at $T$, then $DT\perp EF$.

My Progress: After seeing this problem, the first thing that struck my mind was sharky devil lemma (not a very known lemma )

Here's the lemma: In triangle $ABC$, let $DEF$ be the contact triangle, and let $(M)$ be the midpoint of the arc $(BC)$ not containing $(A)$ in $(ABC)$. Suppose ray $MD$ meets $(ABC)$ again at $R$. If $I$ is the incenter of $(ABC)$ and ray $RI$ intersects $(ABC)$ again at $A'$, then $A'$ is the antipode of $A$. If $P=RA'\cap EF$, then $DP\perp EF$.

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Anyways, here's the problem's diagram:

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Here $J$ is defined as $(ABC)\cap (AEF) .$

Now, if I am able to show that $JITA'$ are collinear, then I am done.

Moreover, I got that $T$ is the radical centre of $(AEF),(GHA')$ and $(AHG)$.

enter image description here

Here, I defined $K$ as $AT\cap (AEF)$.

Now,I thought of using Phantom points . So I defined $T'= \overline{JIA'}\cap EF$ .

We want to show that $T'=T $ . To show that $T'=T $ , we can also show $G,T',H$.

Now, note that $AM\perp EF$.

Let $AJ\cap EF=L$.

So, by radical axis lemma on $(AEF),(ABC) ,(GH'EF)$ , we get $AJ,EF,GA'$ concur at $L$ .

Also we have $T'KMI$ and $AJT'M$ cyclic .

Again by radical axis lemma on $(AEF),(AJT'M),(T'MKI)$ , we get $AJ,TM(EF),KI$ concur at $L$.

Note that $\angle AGA'=90=\angle AMF$ . Since $LFE$ and $LGA$ are collinear , we get $(AHMGLK)$ concyclic.

Also note that $T'$ is the orthocentre of $\Delta ALI$.

This is what I got till now. Now after showing that J,I,T are collinear , by applying sharky devil lemma , we will be done . I know that this problem has a 1 para solution( by @Anand ), but can someone provide a non-projective solution ?

Thanks in Advance!

  • $\begingroup$ not related to the question but do you know some link where they prove the sharky devil lemma? thanks $\endgroup$ Aug 27, 2020 at 4:21
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    $\begingroup$ @hellofriends maybe this might help artofproblemsolving.com/community/c776104h1954097 $\endgroup$ Aug 27, 2020 at 4:32
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    $\begingroup$ I saw the configuration in the Sharky-Devil lemma after a long time, saw it first in a proof of $IO^2=R^2-2Rr$. $\endgroup$ Aug 28, 2020 at 3:22

1 Answer 1


Continuing from where you finished...

Note that $\odot(AEF)$ has $AI$ as diameter. Also, as $T$ is the radical center of $\{\odot(AEF),\odot (GHA'),\odot (AGH)\}$ and thus, $AK\perp LI$ and $LT\perp AI\implies T$ is orthocenter of $\triangle ALI$ and thus, $IT\perp AL$. Let, $IT\cap AL=J'$ and thus, $J'\in\odot(AI)\implies J'=J\implies $ by sharky devil lemma, we get, $J-I-T$ collinear completing the proof.

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    $\begingroup$ Thanks :) . I got it.. $\endgroup$ Aug 27, 2020 at 6:35
  • $\begingroup$ is $K$ defined to be in line $AT$ or $AT'$? Because it seems to me when you prove $T'$ is orthocenter you already have that $T=T'$ $\endgroup$ Aug 29, 2020 at 9:03
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    $\begingroup$ @hellofriends The above proof does not require the definition of $T'$ $\endgroup$
    – Anand
    Aug 29, 2020 at 9:08

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