My proof for that nice problem is rather long - I hope someone will read it and tell me the simple argument afterwards . Let $AI$, $BI$ intersect the circle $\odot(ABC)$ again at $P$, $Q$, respectively. I first prove that the lines $AQ$, $BP$ and $OI$ are concurrent.

Let the common point of $AQ$, $OI$ and $BP$ be $X$ and consider the triangle $\triangle{ABX}$. $BQ$ and $AP$ are altitudes, therefore their point of intersection $I$ is the orthocentre of $\triangle{ABX}$, implying that $XI$ is the third altitude and therefore $XI\perp AB\Leftrightarrow OI\perp AB$.

Notice that $O$ is the midpoint of $A_1B_1$ ,let the foot of the perpendiculars from $A_1,B_1$ to $AB$ be P,Q respectively, it`s easy to prove that $OP=OQ$, and that $IP$ is the external bisector of triangle $\triangle{AB_1P}$ so $\angle{IPQ}=45$ analogously $\angle{IQP}=45$ then $IP=IQ$ so $I$ and $O$ are in the perpendicular bisector of segment $PQ$ and we are done!

pippin 3,14 wrote: [...] and that $IP$ is the external bisector of triangle $\triangle{AB_1P}$ [...] I think you mean $\triangle{AB_1Q}$, but it's just a typo. Nice proof, though and a little shorther than mine xD

Quote: Bisectors $AA_1$ and $BB_1$ of a right triangle $ABC \ (\angle C=90^\circ )$ meet at a point $I.$ Let $O$ be the circumcenter of triangle $CA_1B_1.$ Prove that $OI \perp AB.$ My solution is a little bit complicated. Let $C',B'$ and $A'$ respectively be the tangency points of $(I)$ with $BC,$ $CA$ and $AB$. Denote $M_c$ and $O_c$ by the midpoint of $A'B'$ and $AB$. Note that, since $ABC$ is a $C-$ right triangle, hence $CA'IB'$ is also square $\Longrightarrow$ $M_c$ is the midpoint of $IC$. By Gauss line theorem applying for complete quadrilateral $(BC,BB_1,AC,AA_1)$, we obtain that $M_c,$ $O$ and $O_c$ are collinear; which implies that the circumcircles of $\triangle CA_1B_1,$ $\triangle CAB$ and $IA'CB'$ intersect at two common distinct points: $C$ and $C''$. Now, consider the inversion through pole $I$, power $k=r^2$, where $r$ is the radii of $(I)$; we have $\mathcal {I}(I,k):$ $(O_c)\mapsto$ the Euler circle of $\triangle A'B'C'$, $(M_c)\mapsto A'B'$. Therefore, $C''$ will turn into $H_c$, where $H_c$ is the projection of $C'$ onto $A'B'$. $\mathcal {I}(I,k):$ $C\mapsto M_c;$ $A_1\mapsto A'_1;$ $B_1\mapsto B'_1,$ where $A'_1$ and $B'_1$ are respectively are the projections of $A'$ and $B'$ onto $IA,$ $IB$. As a result, $(O')\equiv (A'_1H_cM_cB'_1)$ will be the image of $(O)$ under $\mathcal {I}(I,k)$. Moreover, $O'$ is also the image of $O$ through the homothety $\mathcal {H} \left (I,\frac {k}{\mathcal {P}_{I/(O)}} \right)$ $\Longrightarrow$ $\overline {I,O,O'}$. We claim that $O'$ is the midpoint of $IC'$. Indeed, let $O'',$ $M_a$ and $M_b$ respectively be the midpoints $IC',$ $B'C',$ $C'A'$. We have $I,M_a$ and $A'_1$ are collinear, so as $I,M_b$ and $B'_1$. Denote $H_b$ by the intersection of $IM_a$ with $C'A'$, $H_a$ by the intersection of $IM_b$ with $C'B'$. Since $\angle A'IB'=90^{\circ}$ $\Longrightarrow$ $\angle A'C'B = 45^{\circ}$, which implies $\triangle C'M_aH_b$ is an isosceles $M_a-$ right triangle. Hence, $\angle M_bH_bI=45^{\circ}$. As a result, $\triangle {M_bIH_b}$ is also an isosceles $M_b-$ right triangle $\Longrightarrow$ $\angle {MIH_b}=45^{\circ}$ $(*)$. Further, since $I,A'_1,A'$ and $M_c$ belong to circle with diameter $IA'$, we have $\angle IA'_1M_c=45^{\circ}$ $(**)$. From $(*)$ and $(**)$, we obtain $IM_b$ $\|$ $M_cA'_1$. Or in other words, $A'_1M_c\| IB'_1$. The same argument for $IA'_1$ and $M_cB'_1$. Therefore, $IA'_1M_cB'_1$ is a parallelogram. Let $M$ be the midpoint of $IM_c$, then $M$ is also the midpoint of $A'_1B'_1$. Let $D'$ be the intersection of $A'_1A'$ with $B'_1B'$ $\Longrightarrow$ $C'A'D'B'$ is a parallelogram, which implies $\overline {C',M_c,D'}$. In the other hand, since $M_cB'_1\| IM_a$ $\perp$ $D'A'_1$, $M_cA'_1\|IM_b$ $\perp$ $D'B'_1$, we obtain that $M_c$ is also the orthocenter of $\triangle D'A'_1B'_1$ $\Longrightarrow$ $D'M_c\equiv C'M_c$ $\perp$ $A'_1B'_1$. Combine with the fact that $O''M$ $\|$ $C'M_c$, we have $O''M\perp A'_1B'_1$ $\Longrightarrow$ $O''A'_1=O''B'_1$. Further, in quadrilateral $C'IM_cH_c$ which is right at vertex $M_c$ and $H_c$, by the property of reflection, it is easy to see that $O''H_c=O''M_c$. In conclusion, $O''$ is the circumcenter of $(A'_1H_cM_cB'_1)$ $\Longrightarrow$ $O''\equiv O'$ $\Longrightarrow$ $\overline {O,I,O',C'}$, which leads to the fact that $OI\perp AB$. Our proof is completed then. $\square$

Dear Mathlinkers, a synthetic proof consists to applied the small Pappus theorem by considering the C-excentre of ABC and then investigate the figure. Sincerely Jean-Louis

Quote: Proposed problem. Let $ABC$ be an $A$-right triangle with incenter $I$ . Denote $\begin{array}{c} E\in BI\cap AC\\\ F\in CI\cap AB\end{array}$ and the midpoint $M$ of $[EF]$ . Prove that $MI\perp BC$ . Proof (synthetic). Denote the projections $X$ , $Y$ on $BC$ of $F$ , $E$ respectively. Observe that $EA=EY$ and $FA=FX$ . Therefore, $IA=IY$ and $IA=IX$ , i.e. $I$ is the circumcenter of $\triangle XAY$ . In conclusion, $MI$ is middleline in the right trapezoid $EFXY$ $\implies$ $MI\perp BC$ . Remark. Prove easily that the point $I$ is the $B$-exincenter of $\triangle BFX$ and is the $C$-exincenter of $\triangle CEY$ . Thus $IX\perp IY$ and $m\left(\widehat{XAY}\right)=45^{\circ}$ .

Dear Virgil and Mathlinkers, your proof is very elegant. Sincerely Jean-Louis

Note the simple fact, $ XY $ is the radical axis of $ \odot CA_1B_1 $ and the incircle of $ \Delta ABC $, where $ X,Y $ are the midpoints of $ CA $ and $ CB $ respectively. This follows the wanted result.

Debdyutida,can you plz explain why XY is the radical axis??

RSM wrote: Note the simple fact, $ XY $ is the radical axis of $ \odot CA_1B_1 $ and the incircle of $ \Delta ABC $, where $ X,Y $ are the midpoints of $ CA_1 $ and $ CB_1 $ respectively. This follows the wanted result. I can be wrong, but if the incircle touch $BC$ at $R$, then $XR^2=XF*XC=XF^2$ but $R$ is between $x$ and $F$, so it hasn't a same power.

Babai wrote: Debdyutida,can you plz explain why XY is the radical axis?? $ XB_1.XC=(\frac {b}{2})(\frac {b}{2}-\frac {ba}{a+c})=\frac {b^2(c-a)}{4(c+a)}=\frac {b^2(c-a)^2}{4(c^2-a^2}=\frac {(c-a)^2}{4} $ Suppose, incircle of $ ABC $ touches $ AC $ at $ T $, $ XT^2=(\frac {b-(b+a-c)}{2})^2=\frac {(a-c)^2}{4} $ So $ XB_1.XC=XT^2 $, similar for $ Y $, so the result follows. @SCP,there was a typo in my post, I have edited it.

From $\angle AIB_1=\angle ICB_1=\angle A_1IB=\angle ICA_1=45^\circ$ we get $AI$ tangent to the circle $\odot CIB_1$ and $BI$ tangent to the circle $\odot ICA_1$, or $AI^2=AB_1\cdot AC$ and $BI^2=BA_1\cdot BC$. That means $I$ and circumcenter of $\triangle B_1CA_1$ are on the same perpendicular to $AB$ (Carnot lemma). Best regards sunken rock

sunken rock wrote: That means $I$ and circumcenter of $\triangle B_1CA_1$ are on the same perpendicular to $AB$ (Carnot lemma). Best regards sunken rock What is carnot lemma?

$AO^2-AI^2=BO^2-BI^2 \iff OI\perp AB$. Best regards, sunken rock

Proof 1 (metric). I"ll use the equivalence $MI\perp BC\ \iff\ MB^2-MC^2=IB^2-IC^2$ and the well-known relations $\{\begin{array}{c} BE^2=ac-EA\cdot EC\\\ CF^2=ab-FA\cdot FB\end{array}\|\ (1)$ . Observe that $\{\begin{array}{ccc} BF=\frac {ac}{a+b}=\frac {a(a-b)}{c}\\\ CE=\frac {ab}{a+c}=\frac {a(a-c)}{b}\end{array}\implies$ $\boxed{c\cdot BF-b\cdot CE=a(c-b)}\ (2)$ . Apply the theorem of the median to : $\{\begin{array}{ccc} BM/\triangle EBF & \implies & 4\cdot BM^2=(BF^2+BE^2)-EF^2\\\ CM/\triangle ECF & \implies & 4\cdot CM^2=(CE^2+CF^2)-EF^2\end{array}$ $\implies$ $2\cdot(MB^2-MC^2)=$ $(BF^2-CE^2)+BE^2-CF^2\stackrel{(1)}{=}$ $(BF^2-CE^2)+(ac-EA\cdot EC)-(ab-FA\cdot FB)=$ $a(c-b)+BF(BF+FA)-CE(CE+EA)=$ $a(c-b)+c\cdot BF-b\cdot CE\stackrel{(2)}{=}$ $2a(c-b)\implies$ $\boxed{MB^2-MC^2=a(c-b)}\ (3)$ . On other hand, $IB^2-IC^2=\frac {ac(s-b)}{s}-\frac {ab(s-c)}{s}\implies$ $\boxed{IB^2-IC^2=a(c-b)}\ (4)$ . From the last relations $(3)$ and $(4)$ obtain that $MB^2-MC^2=IB^2-IC^2$ , i.e. $MI\perp BC$ . Proof 2 (metric). Denote the point $N\in EF$ for which $NI\perp BC$ . I"ll show that $N\equiv M$ . Using an well-known property $\frac {NF}{NE}=\frac {IF}{IE}\cdot\frac {\sin\widehat{NIF}}{\sin\widehat{NIE}}=$ $\frac cb\cdot\frac {CF}{BE}\cdot\frac {\cos\frac C2}{\cos\frac B2}=$ $\frac {a+c}{a+b}\cdot \frac {\cos^2\frac C2}{\cos^2\frac B2}=$ $\frac {c(a+c)(a+b-c)}{b(a+b)(a+c-b)}=$ $\frac {bc(a+c)+c(a^2-c^2)}{bc(a+b)+b(a^2-b^2)}=$ $\frac {bc(a+c)+b^2c}{bc(a+b)+bc^2}=1\implies$ $NE=NF\implies N\equiv M$ . Remark. Prove easily that in any triangle $ABC$ exists the relation $MB^2-MC^2=a(c-b)\cdot[1+\frac {bc\cdot \cos A}{(a+b)(a+c)}]=a(c-b)\cdot\frac {(a+b+c)^2}{2(a+b)(a+c)}$ . Observe that $A=90^{\circ}\iff b^2+c^2=a^2$ $\iff (a+b+c)^2=$ $2(a+b)(a+c)\iff$ $c(a+c)(a+b-c)=b(a+b)(a+c-b)\iff MI\perp BC\iff NE=NF$ .

sunken rock wrote: From $\angle AIB_1=\angle ICB_1=\angle A_1IB=\angle ICA_1=45^\circ$ we get $AI$ tangent to the circle $\odot CIB_1$ and $BI$ tangent to the circle $\odot ICA_1$, or $AI^2=AB_1\cdot AC$ and $BI^2=BA_1\cdot BC$. That means $I$ and circumcenter of $\triangle B_1CA_1$ are on the same perpendicular to $AB$ (Carnot lemma). $BO^2=OA_1^2+BA_1^2+2A_1B\cdot A_1O \cdot cos(OA_1C)=BO^2=OA_1^2+BA_1^2+CA_1\cdot BA_1 \\\\ AO^2=OB_1^2+AB_1^2+2B_1A\cdot B_1O \cdot cos(OB_1C)=CO^2=OB_1^2+AB_1^2+B_1C\cdot B_1A $ Hence $BO^2-AO^2=BI^2-AI^2=BA_1\cdot BC-AB_1\cdot AC$

Let C(0,0), A(b,0), B (0,a), I(Ix, Iy), O(Ox, Oy) $ a^2+b^2=1$ => $Ix=Iy=\frac {ab}{1+a+b}$ $Ox=\frac {ab}{2+2a}, Oy=\frac {ab} {2+2b}$ => Slope of OI $= \frac {(1+a)(1+b-a)}{(1+b)(1+a-b)} =\frac {1-a^2+ab+b} {1-b^2+ab+a} =\frac {b^2+ab+b}{a^2+ab+a}=\frac {b}{a}$

Let $A_1X$ and $B_1Y$ be perpendicular to $AB$. we will prove $IX^2 - IY^2 = OX^2 - OY^2$. Note that $A_1XYB_1$ is trapezoid and $O$ is midpoint of $A_1B_1$ so $OX = OY$ so we need to prove $IX = IY$. Note that $I$ lies on angle bisector of $\angle A$ and $\angle YB_1C$ so $\angle IYX = \angle 45$. with same approach we have $\angle IXY = \angle 45$ so $IX = IY$.