Dear Mathlinkers, in order to make a link http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=358741. Sincerely Jean-Louis

If $X \in BC, Y \in CA, Z \in AB$ are feet of the external angle bisectors of $\triangle ABC$ $\Longrightarrow$ $DE \parallel XYZ \perp OI.$ For example, see http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=202692 and elsewhere. WLOG, $AB > BC.$ Parallel to $BC$ through $D$ cuts $AC$ at $F.$ $R_A, R_F$ are circumradii of $\triangle ADE, \triangle FDE.$ $\frac{R_A}{R_F} = \frac{AD}{DF} = \frac{AB}{BC}$ and $R_A - R_F = \frac{AD}{AB} \cdot OI$ $\Longrightarrow$ $\frac{AB - BC}{AB} \cdot R_A = \frac{AD}{AB} \cdot OI$ $\Longrightarrow$ $R_A = OI.$ See http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=83383 for both synthetic and trigonometric proofs for the $R_A - R_F$ value. Note that the trigonometric proof is almost trivial. More trigonometric proofs for the $R_A$ value are at http://www.artofproblemsolving.com/Forum/viewtopic.php?t=14655.

Dear Mathlinkers, an article concerning some results about “two equal segments on two sides of a triangle” has been put on my website http://perso.orange.fr/jl.ayme vol. 6 Deux segments égaux sur deux côtés d'un triangle p. 25 You can use Google translator Sincerely Jean-Louis

Let $M$ and $K$ be midpoint of $AB$ and $AC$, respectively. Let incircle of $\triangle ABC$ touch $AB$ and $AC$ at $N$ and $L$, respectively. $BM=\frac c2$, $BN=u-b$. So $MN=\frac{b-a}{2}$. Similary, $KL=\frac{c-a}2$. Let the circle with diameter $OI$ cut $IN$ at $R$, $OK$ at $S$. Let $T$ be a point on that circle with $OS \parallel RT$. We have $ST=OR=\frac {b-a}2 = \frac{AE}2$ and $SI=\frac{c-a}2 = \frac{AD}2$. Since $OS \parallel RT \Rightarrow SI \perp RT$, so $\angle TRI + \angle RIS = \angle TSI + \angle RIS= 90^\circ \Rightarrow \angle RIS = 90^\circ - \angle TSI$. In $ANIL$, $\angle BAC + \angle NIL = 180^\circ \Rightarrow \angle RIS = \angle NIL - 90^\circ = 90^\circ - \angle BAC$. So $\angle BAC = \angle TSI$. Also we have $\frac {ST}{AE} = \frac {SI}{AD} = \frac 12$. By $S.A.S$, $\triangle TSI \sim \triangle EAD$ with ratio $\frac 12$. This concludes circumradius of $\triangle ADE$ is twice of circumradius of $\triangle TSI$ which is the diameter of circumcircle of $\triangle TSI$. This is nothing but $OI$. $\blacksquare$ As a bonus, I will prove $OI \perp DE$. As many people mention, this is a famuous problem asking for both $OI\perp DE$ and $OI=DE$. The circle with diameter $OI$ gives $\angle ADE = \angle TIS = \angle OIR$. So $OI \perp DE$.

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Let $BI\cap (ABC)=K$ and $CI\cap (ABC)=L$. Let $O_1$ be the center of $(ADE)$. Since $|BC|=|BD|$, one can find that $\angle CIA=90^\circ+\frac{\angle ABC}2=180^\circ-\left(90^\circ-\frac{\angle ABC}2\right)=\angle CDA\Rightarrow ACID$ is cyclic. We know that $K$ is the center of $(ACID)$. Similarly, $ABIE$ is cyclic and $L$ is the center of this circle. We know that the reflection of $A$ with respect to $KL$ is $I$.......(1) Since $|O_1A|=|O_1D|$ and $|KA|=|KD|$, we find that $KO_1\perp AB$. Also, $OL\perp AB$. Hence, $KO_1//OL$. Similarly, $LO_1//OK$. Thus, $OKO_1L$ is a parallelogram. Also, $|OK|=|OL|$. Hence, $OKO_1L$ is a rhombus. Therefore, the reflection of $O_1$ with respect to $KL$ is $O$.......(2) Now, combine $(1)$ and $(2)$ This gives us that $AO_1OI$ is an isosceles trapezoid. Hence, $R_{ADE}=|O_1A|=|OI|$, as desired.