I have solved it using the formula d^2=R(R-2r) and so use sine law to deduce the two lines are perpendicular. However, I think it is too troublesome and I believe that there is a simpler proof. Would someone work it out?

I think this was a Crux problem some years ago. I will prove a more general result: Let ABC be an arbitrary triangle, and D and E two points on the halflines AB and CB such that AD = CE = CA. Prove that $OI\perp DE$ and $DE=2\sin B\cdot OI$, where O is the circumcenter and I is the incenter of triangle ABC. Here is a quite smart proof of this (not from me; Peter surely will remember my loooong proof using vectors, but this one is much simpler): We will assume that b = CA is the shortest side of the triangle ABC, but in all other cases the proof will work as well, with some slight modifications. The circumcenter O of the triangle ABC lies on the perpendicular bisectors of the segments BC, CA, AB. Let D' and E' be the orthogonal projections of the point I on the perpendicular bisectors of the segments BC and AB. Also, let A' and C' be the midpoints of the segments BC and AB; and let X and Z be the orthogonal projections of the point I on the lines BC and AB, i. e. the points where the incircle of triangle ABC touches the sides BC and AB. It is well-known that BX = s - b, where $s=\frac{1}{2}\left( a+b+c\right)$ is the semiperimeter of the triangle ABC. Thus, $XA^{\prime }=BX-BA^{\prime }=\left( s-b\right) -\frac{1}{2}a=\left( \frac{1}{2}\left( a+b+c\right) -b\right) -\frac{1}{2}a=\frac{1}{2}\left( c-b\right) $. Since the quadrilateral IXA'D' is a rectangle (it has three right angles: at X, at A' and at D'), we have ID' = XA', so $ID^{\prime }=\frac{1}{2}\left( c-b\right) $. Now we have BD = AB - AD = AB - CA = c - b; therefore, $ID^{\prime }=\frac{1}{2}\left( c-b\right) =\frac{1}{2}\cdot BD$. Similarly, $IE^{\prime }=\frac{1}{2}\cdot BE$. Finally, we have $D^{\prime }I\perp A^{\prime }D^{\prime }$ and $A^{\prime }D^{\prime }\perp BC$, i. e. $A^{\prime }D^{\prime }\perp EB$, so that D'I || EB. Similarly, E'I || DB. Hence, the angle < D'IE' is equal to < DBE, since the sides of the two angles are respectively parallel. Together with $ID^{\prime }=\frac{1}{2}\cdot BD$ and $IE^{\prime }=\frac{1}{2}\cdot BE$, this shows that the triangles DBE and D'IE' are similar, with the ratio of similitude $\frac{1}{2}$. Hence, $D^{\prime }E^{\prime }=\frac{1}{2}\cdot DE$. Now, since < OD'I = 90° and < OE'I = 90°, the points D' and E' lie on the circle with diameter OI. In other words, the circumcircle of triangle D'IE' has the segment OI as diameter. Therefore, after the extended law of sines, $D^{\prime }E^{\prime }=OI\cdot \sin \measuredangle D^{\prime }IE^{\prime }$. Hence we have $\frac{1}{2}\cdot DE=D^{\prime }E^{\prime }=OI\cdot \sin \measuredangle D^{\prime }IE^{\prime }=OI\cdot \sin \measuredangle DBE=OI\cdot \sin B$, and thus $DE=2\cdot OI\cdot \sin B=2\sin B\cdot OI$. This already proves one part of our assertion. Remains to show $OI\perp DE$. Now, we will work with directed angles modulo 180°. Actually, the triangles DBE and D'IE' are not just similar, they are inversely similar (since the angles < DBE and < D'IE' are oppositely equal, if we consider them as directed angles). Hence, < ID'E' = - < BDE. But since the points D' and E' lie on the circle with diameter OI, we have < IOE' = < ID'E', so < IOE' = - < BDE. Hence, < (OI; OE') = < IOE' = - < BDE = < EDB = < (DE; AB), and thus < (OI; DE) = < (OI; OE') + < (OE'; AB) + < (AB; DE) = < (OI; OE') + < (OE'; AB) - < (DE; AB) = < (OE'; AB) = 90° (since $OE^{\prime }\perp AB$). Therefore, $OI\perp DE$, qed.. Darij

Note that in the case where <B=30 (not in the general case!), we can also prove that I is the orthocentre of ODE. This case is much easier than the general case, mainly because here we have OAC is equilatrel. First, the triangles EIC, AIC and AID are congruent. Since <AIC=105, then <AID=<CIE=105 and <DIE=45. For proving IO is perpendicular to DE, it is sufficient to prove OE is perpendicular to ID, i.e. <OEI=45. Since O is the circumcentre of ABC, <AOC=2*<ABC=60 and AOC is equilatrel and AO=AC=OC=EC=DA so OEC is iscoceles. <OCE=<OCB=90-A, so <CEO=(90+A)/2. But <IEC=<IAC=A/2, so <IEO=(90+A)/2 - A/2=45. So it follows O is the orthocentre of IDE. Proving IO=DE is equivalant to the following result: Let XYZ be a triangle, H its orthocentre, with <YXZ=45. Then XH=YZ. This is easy since HA=a/tanA.

I just got a short proof of the indentity darij grinberg posted. Let $AI$ intersects the circumcircle of $ABC$ at $A'$ and AA' cuts CD at F. It's obvious that $OA'$ is perpendicular to $BC$ and AA' is perpendicular to CD. So $<OA'B=<OA'C=<BA'C/2=(180-<BAC)/2=<ADC=<ACD$. Hence $<OA'I=<OA'B-<IA'B=<ACD-<ACB=ECD$ --- (1) Also since $\frac{EC}{OA'}=\frac{b}{R}=2\sin B=2\sin <FA'C=2\frac{CF}{CA'}=\frac{CD}{A'I}$ --- (2) , from (1) and (2) we get triangle $ECD$ is similar to triangle $OA'I$. Since $OA'$ is perpendicular $EC$, so $DE$ is perpendicular to $OI$ and $DE=2\sin B\cdot OI$.

darij grinberg wrote: I think this was a Crux problem some years ago. I will prove a more general result: $ XA^{\prime }=BX-BA^{\prime }=\left( s-b\right)-\frac{1}{2}a=\left( \frac{1}{2}\left( a+b+c\right)-b\right)-\frac{1}{2}a=\frac{1}{2}\left( c-b\right)$. Does $ XA^{\prime }=BX-BA^{\prime }$ always true?

Leonhard Euler wrote: darij grinberg wrote: I think this was a Crux problem some years ago. I will prove a more general result: $ XA^{\prime }=BX-BA^{\prime }=\left( s-b\right)-\frac{1}{2}a=\left( \frac{1}{2}\left( a+b+c\right)-b\right)-\frac{1}{2}a=\frac{1}{2}\left( c-b\right)$. Does $ XA^{\prime }=BX-BA^{\prime }$ always true? Actually, it's rather $ XA^{\prime }=\left|BX-BA^{\prime }\right|$. I wrote the above referring to my sketch, where $ BX\geq BA^{\prime}$. If $ BX\leq BA^{\prime}$, then the proof needs some slight modifications (but is analogous). One day I will learn Asymptote and start attaching graphics to my posts... Darij

Enunciation. Let $ ABC$ be a triangle with the circumcircle $ w=C(O,R)$ and the incircle $ C(I,r)$. Define the points $ \{\begin{array}{c}D\in (AB\ ,\ E\in (CB\\\ AD = CE = CA\end{array}$. Prove that $ OI\perp DE$. Proof. Denote $ p_{w}(X)$- the power of the point $ X$ w.r.t. the circle $ w$. Thus, $ \{\begin{array}{c}p_{w}(D)\equiv OD^{2}-R^{2}=-b(c-b)\\\ p_{w}(E)\equiv OE^{2}-R^{2}=-b(a-b)\end{array}\|$ $ \implies$ $ \boxed{OD^{2}-OE^{2}=b(a-c)}$. The relations $ IA^{2}=\frac{bc(p-a)}{p}$ a.s.o. are well-known. Apply the Stewart's theorem to the rays $ (ID$, $ (IE$ in $ \triangle IAB$, $ \triangle IAC$ respectively : $ \{\begin{array}{c}b\cdot IB^{2}+(c-b)\cdot IA^{2}=c\cdot ID^{2}+bc(c-b)\\\ b\cdot IB^{2}+(a-b)\cdot IC^{2}=a\cdot IE^{2}+ab(a-b)\end{array}\|$ $ \implies$ $ \boxed{ID^{2}-IE^{2}=b(a-c)}$. In conclusion, $ OD^{2}-OE^{2}=ID^{2}-IE^{2}$, i.e. $ OI\perp DE$.