See the diagram for a proof without words.

Probably not the intended solution, but a solution nevertheless. Let $M=AX\cap BY$, $N=BY\cap CZ$, $P=CZ\cap AX$. Note that $AX$ is the radical axis of $(AEF)$ and $(ABC)$ and that $BY$ is the radical axis of $(BED)$ and $(ABC)$, so $M$ lies on the radical axis of $(AEF)$ and $(BED)$. However, since $AE\perp EI$ and $AF\perp FI$, where $I$ is the incenter, $AEIF$ is a cyclic quadrilateral, so $M\in EF$ and $ME\perp AD$. Similarly, $ND\perp BC$ and $PF\perp CA$. Therefore \begin{align*}AM^2+BN^2+CP^2&=(AE^2+EM^2)+(BD^2+DN^2)+(CF^2+FP^2)\\&=(BE^2+EM^2)+(CD^2+DN^2)+(AF^2+FP^2)=MB^2+NC^2+PA^2,\end{align*} so by Carnot's Theorem the perpendiculars to $MN, NP, PM$ from $B,C,A$ respectively are concurrent as desired. $\blacksquare$

Dear Mathlinkers, after the nice and silent proof of Luis, who want to give the geometric nature of P? Sincerely Jean-Louis

Dear Mathlinkers, P is the Bevan's point of ABC... see http://perso.orange.fr/jl.ayme vol. 3, Cinq théorèmes de von Nagel p. 24-25 Sincerely Jean-Louis

Isn't this problem true even if $I$ is not the incenter?

Luis González wrote: See the diagram for a proof without words. Nice!

jelena_ivanchic wrote: Let $I,O$ be incenter and circumcenter of $\triangle ABC$. Its well known that $O$ is the midpoint of $IV$. Now, let $A'$ be antipode of $A$ wrt $\odot(O)$. Obviously, $\triangle OIA'\equiv \triangle OVA$ and thus, $IA'\|AV$ and we are done by sharky-devil lemma. Now applying sharky-devil lemma to every vertex, the claim follows as they concur at Bevan point($V$)

We claim that the reflection of $I$ in $O$ defined as $V$ is the desired point of concurrency . Note that that since $\angle {AXI} =90^ {\circ}$ , we need to show $AV \parallel IA'$ .(Easy to see $X-I-A'$) Where $A'$ is the $A$ antipode of $\odot (ABC)$ . This is just straight forward complex numbers with $\odot(ABC)$ as the unit circle and $A=a^2$ ,etc Note that $$AV \parallel IA' \iff \frac {v-a}{i-a'} \in \mathbb {R} \iff \frac {ab+bc+ca-a^2}{-ab-bc-ca+a^2} \in \mathbb {R} \iff -1 \in \mathbb {R} $$ Hence $AV \perp AX$ and we're done by symmetry . $\blacksquare$ Taha1381 wrote: Isn't this problem true even if $I$ is not the incenter? Yes I believe . The complex number can approach can be adapted suitably

Lemma[Six point circle theorem] Let $ABC$ be a triangle and let $X'$ be isogonal conjugate of $X$ wrt $ABC$. Then the feet of $X$ and $X'$ on $AB,BC,CA$ form a cyclic hexagon. Proof. Let feet from $X$ to $BC,AC,AB$ are $D,E,F$ respectively and feet from $X'$ to $BC,AC,AB$ are $D',E',F'$. We have \begin{align*} \measuredangle DEE'&=\measuredangle DEC\\ &=\measuredangle DXC\\ &=90^\circ -\measuredangle XCD\\ &=\measuredangle DD'X+\measuredangle DCX\\ &=\measuredangle DD'X+\measuredangle X'CA\\ &=\measuredangle DD'X+\measuredangle X'D'E'\\ &=\measuredangle DD'E, \end{align*}and thus $DD'EE'$ is cyclic. Similarly we obtain that $EE'FF'$ and $DD'FF'$ are cyclic. Suppose those circles are pairwise different, then by radical axis, we should have $AB,AC,BC$ concurrent, which is clearly false, thus in fact $DD'EE'FF'$ is cyclic hexagon. $\square$ Let $P=AX\cap BY$, $Q=AX\cap CZ$ and $R=BY\cap CZ$. Let $K$ be the isogonal conjugate of $I$ wrt $PQR$. Let $A',B',C'$ be the feet from $K$ on $PQ,PR,RQ$. Then by the lemma above, we obtain that $A'B'C'XYZ$ is cyclic hexagon, however $ABCXYZ$ is also cyclic. Hence, $A'=(XYZ)\cap PQ=A$, $B'=(XYZ)\cap PR=B$ and $C'=(XYZ)\cap QR=C$, we get desired concurrency. $\blacksquare$

Let the aforementioned perpendiculars from $A, B, C$ meet $(ABC)$ again at $A_1, B_1, C_1$ respectively, the antipodes of $A, B, C$ wrt $(ABC)$ be $A', B', C'$ respectively, and $O$ be the circumcenter of $ABC$. It's easy to see $AA_1A'X$, $BB_1B'Y$, and $CC_1C'Z$ are rectangles. Since $$\angle AXA' = 90^{\circ} = \angle AXI$$we know $XA'$ passes though $I$. Via analogous processes, we conclude $XA', YB', ZC'$ are concurrent at $I$. Now, consider the homothety at $O$ with scale factor $-1$. Notice $XA', YB', ZC'$ map to $A_1A, B_1B, C_1C$ respectively. As a result, we know $AA_1, BB_1, CC_1$ concur at the image of $I$, as required. $\blacksquare$ Remark: See my GeoGebra diagram here!

Let AX&BY intersect at C';BY&CZ intersect at A';CZ&AX intersect at B'.Let D, E, F be the tangency points of the incircle with BC, CA, AB respectively. By radical axis theorem on (BFID), (AFIE)&(ABC);C'-F-I are collinear. Similarly B'-E-I are collinear & A'-D-I are collinear. Since ID,IE, IF are perpendicular to BC,CA,AB respectively, the triangles ABC & A'B'C' are orthologic which implies the result.

Let $P=\overline{BY}\cap\overline{CZ}\cap\overline{ID},Q=\overline{CZ}\cap\overline{AX}\cap\overline{IE},$ and $R=\overline{AX}\cap\overline{BY}\cap\overline{IF}$ by Radical Axis. Then, \begin{align*}(AQ^2-AR^2)+(BR^2-BP^2)+(CP^2-CQ^2)&=EA^2+EQ^2-FA^2-FR^2\\&+FB^2+FR^2-DB^2-DP^2\\&+CE^2+EP^2-EA^2-EP^2\\&=0\end{align*}and we are done by Carnot's Theorem. $\square$

Let $V$ be the reflection of $I$ across $O$ (i.e. the Bevan point of $\Delta ABC$). We show that $V$ is the desired concurrency point. Let $A',X'$ denote the antipodes of $A$ and $X$ in $\Gamma$ respectively. Note that $AI$ is the diameter of $(AEF)$, so $X,I,A'$ are collinear. By taking a homothety about $O$ with scale factor $-1$, we see that $A,V,X'$ are collinear, so $\angle XAV = \angle XAX' = 90^\circ$. Hence proved.