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Extend side $AC$ of triangle $ABC$ to point $E$ such that $AB=AE$.

Extend side $BC$ of triangle $ABC$ to point $D$ such that $AB=BD$.

Denote the center of circumscribed circle of triangle $ABC$ with $O$ and center of inscribed circle of the same triangle with $I$.

Prove that: $OI\perp ED$.

I have tried to translate the problem into the complex plane but the calculations became messy fairly quickly. This is a contest math problem, one of those than don't give you much clues on how to proceed. If you draw too many lines, just to outline the position of $O$ and $I$ the picture becomes messy too.

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    $\begingroup$ A simple solution comes from an obscure lemma: for any sufficiently large constant $k$, the locus of points $P$ such that $d(P,AB)+d(P,AC)+d(P,BC)=k$ is a line orthogonal to the $OI$-line. This can be proved through Lagrange multipliers and/or trilinear coordinates, for instance. $\endgroup$ – Jack D'Aurizio Sep 17 '18 at 11:33
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As I don't know any synthetic geometry trick to solve this problem, my approach to tackle it is an analytical one.

I'm going to calculate the equation of line DE and the equations of circumcircle and incircle of triangle ABC. From the equations of these circles I will get the equation of the radical axis, and then it will be easy to show that OI is perpendicular to line DE

Let's begin:

Let the point C be the origin of cartesian oblique axes, so that line ACE becomes the x-axis and line BCD becomes the y-axis. Then the coordinates of C are $C=(0,0)$ and the coordinates of B and A can be, without loss of generality, $B=(0,q)$ and $A=(p,0)$, with $p<0$ and $q<0$. Then, if we call c the measure of side AB, the coordinates of D and E are $D=(0,q+c)$ and $E=(p+c,0)$.

With these coordinates, we immediately note that line DE has the equation $$(q+c)x+(p+c)y-(q+c)(p+c)=0$$

The equation of the circumcircle, which passes through points A, B, C, is obviously

$$x^2+2xy\cos C +y^2 -px -qy=0$$

As for the incircle, it's not hard to see that it touches the x-axis and y-axis in the points $((p+q+c)/2,0)$ and $(0,(p+q+c)/2)$, which leads us to almost effortlessly arrive at its equation: $$x^2+2xy\cos C +y^2 -(p+q+c)x -(p+q+c)y +(1/4)(p+q+c)^2=0$$

Subtracting both equations of these circles we finally get the equation of their radical axis:

$$(q+c)x+(p+c)y-(1/4)(p+q+c)^2=0$$

Comparing it with the equation of DE we have previously got, we see that both are parallel to each other, so that OI is perpendicular to DE.

Solved.

This problem, by the way, shows us two important points:

i) we shouldn't disparage or forget the cartesian oblique axes;

ii) how a wise choice of a coordinate system, axes and origin can pave the way to a smooth solution.

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A second solution, using complex numbers:

Let the circumcenter O be the origin and let us assign the complex numbers $z_1$, $z_2$, $z_3$, $z_4$, $z_5$, $u$ to the points A, B, C, D, E, I.

As $\angle IAB=\angle EAI=A/2$, $AE=AB$, and $AI$ is a common side, $\triangle AIE \cong \triangle AIB$ (SAS criterion). Thus $IE=IB$ and $\angle AEI=\angle IBA=B/2$.

Therefore the point E can be thought as the image of a rotation of point B about the incenter I through an angle equal to $(A+B)$:

$$z_5-u=(z_2-u)e^{i(A+B)}$$

Likewise the point D can be thought as the image of a rotation of point A about the incenter I through an angle equal to $-(A+B)$:

$$z_4-u=(z_1-u)e^{i(-(A+B))}$$

Subtracting both equations we get:

$$z_4-z_5=z_1.e^{i(-(A+B))} -z_2.e^{i(A+B)}+u(2i\sin(A+B)),$$ $$z_4-z_5=z_1.e^{i(C-\pi)} -z_2.e^{i(\pi-C)}+u(2i\sin C),$$

On the other hand, as $z_2=z_1.e^{i(2C)}$,

$$z_4-z_5=z_1.e^{i(C-\pi)} -z_1.e^{i(2C)}.e^{i(\pi-C)}+u(2i\sin C),$$

$$z_4-z_5=z_1.e^{i(C-\pi)} -z_1.e^{i(\pi+C)}+u(2i\sin C),$$

$$z_4-z_5=u(2i\sin C),$$

$${z_4-z_5\over u}= 2i\sin C$$

Therefore, as $\sin C \neq 0$ and $\sin C$ is a real number,

$$OI \perp DE$$

QED.

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