Reflect $D$ over $B$ to get $D'$ and $E$ over $C$ to get $E'$ it's sufficient to prove that $\angle E'I_aD'=\pi-\alpha$.Let $\odot I_a$ touch $AB,AC$ in $P,Q$.$D'P=E'Q$ $\implies$ $\triangle I_aD'P\cong \triangle I_aE'Q$ $\implies$ $\angle E'I_aD'=\angle PI_aQ=\pi-\alpha$ $\blacksquare$

Reflect $D$ over $B$ to get $D'$ and $E$ over $C$ to get $E'$ Let $K$ be the intersection of $(ADE)$ and $(ABC)$. It's trivial that $\triangle KBD' \thicksim \triangle KIJ \thicksim \triangle KCE'$ and thus $A,D',J,E',K$ are concyclic,as desird.

j___d wrote: Incircle of a triangle $ABC$ touches $AB$ and $AC$ at $D$ and $E$, respectively. Point $J$ is the excenter of $A$. Points $M$ and $N$ are midpoints of $JD$ and $JE$. Lines $BM$ and $CN$ cross at point $P$. Prove that $P$ lies on the circumcircle of $ABC$. My solution: Let $I$ be the incenter and let $T$ be the midpoint of $JI$,it is well known that $T$ lies on the circumcircle of $ABC$ so it's sufficient to prove that $BCTP$ is cyclic Note that $NT,MT$ are parallel to $EI,DI$ respectively so $\angle MTN$=$\angle BTC$=$180-\angle A$ so $\angle NTC$=$\angle BTM$ and by $TN=TM$ and $BT=TC$ we get that $\triangle CNT$=$\triangle TMB$ so $\angle NCT$=$\angle MBT$ and so done.

We use barycentric coordinates with respect to triangle $ABC$. Denote $s=\frac{a+b+c}{2}$. We can simply check that $J=(\frac{-a}{2s-2a}; \frac{b}{2s-2a};\frac{c}{2s-2a})$, $M=(\frac{-a}{2s-2a}+\frac{s-b}{c}:\frac{b}{2s-2a}+\frac{s-a}{c}: \frac{c}{2s-2a})$ and $N=(\frac{-a}{2s-2a}+\frac{s-c}{b}:\frac{b}{2s-2a}:\frac{s-a}{b}+\frac{c}{2s-2a})$. Let $m=2(s-b)(s-a)-ac$ and $n=2(s-c)(s-a)-ab$. Then $P=(mn: mb^2:nc^2)$. So we need check if: $$(abc)^2mn+(bc)^2mn^2+(bc)^2nm^2=0$$. This is equivalent to: $$a^2+m+n=a^2+2(s-a)(2s-b-c)-a(b+c)=a^2+a(2s-2a-b-c)=a^2-a^2=0$$We are done.

Let $a,b,c$ be complex numbers lying on unit circle centered at $0$ such that $a^2,b^2,c^2$ are vertices of the triangle and $-(ab+bc+ca)$ is the incenter. These numbers always exist. Then $$2d=a^2+b^2-ab-bc-ca+ab\cdot\frac{a+b+c}{c},\ e=ab-bc+ac,\ 4m=2d+2e=a^2+b^2+ab+ac-3bc+ab\cdot\frac{a+b+c}{c}$$Let $Q\neq B$ be the intersection of line $BM$ with circumcircle of $ABC$. Then $q\cdot\overline{q}=1$ and $$Q\in BM\iff \frac{b^2-q}{b^2-m}=\overline{\left(\frac{b^2-q}{b^2-m}\right)}\iff q=\frac{b^2-m}{b^2\cdot\overline{m}-1}$$Since $$4b^2-4m=(b+c)\cdot\frac{-a(a+b+c)+3bc}{c},\ 4(b^2\cdot\overline{m}-1)=(b+c)\cdot\frac{ab+bc+ca-3a^2}{a^2c}$$we have$$q=\frac{-a(a+b+c)+3bc}{ab+bc+ca-3a^2}$$Observe this is symmetric in $(b,c)$ so the same are coordinates for intersection of $CN$ with circumcircle of $ABC$ other than $C$. Thus $P=F$ qed

easy and nice problem Claim : BM=CN Proof: 2.$BM^2$=$BJ^2 $+ $BD^2 $- 1/2.$JD^2 $and 2.$CN^2$=$ CE^2 $+$ CJ^2 $- 1/2. $JE^2$ since JE=JD(easy to prove) it is enough to prove that $BD^2$-$CE^2 $+$ BJ^2$- $CJ^2$=0 we know that $BD^2$- $CE^2$= $BI^2 $- $ID^2$-$IC^2$+$IE^2 $=$IB^2 $- $IC^2 $since $\angle IBJ$ = $\angle ICJ$ = 90 proof finishes Proof of problem : B' is the reflection of B over AJ. B' lie on line AC. the reflection of P over AJ is P' and the reflection of M over AJ is N. it is enough to prove that $\angle AB'P'$ + $\angle ACP$ =180 B' and N and P' are colinear now it is enough to prove that B'N=CN=BM and that is true and proof is completed now

Let $S$ and $K$ be reflections of $D$ and $E$ across $B$ and $C$. Note that $ BP || SJ$ and $CP || KJ$ so we will prove $ASJK$ is cyclic. Let incircle of $ABC$ meet $BC$ at $F$. claim1 : $BSJ$ and $BFJ$, $CFJ$ and $CKJ$ are congruent. Proof : $BS = BD = BF$, $BJ = BJ$ and $\angle JBS = \angle KBF$ so $BSJ$ and $BFJ$ are congruent. we prove the other one with same approach. Now we have $\angle SJK = 2\angle BJC = 180 - \angle A$ so $ASJK$ is cyclic. we're Done.

Oooh Nice Problem! No one has posted my solution yet (Of course considering my obsession with miquel points ) Let $L$ be the midpoint of minor arc $\overarc{AB}$ Notice because of IEL $L$ is the midpoint of $IJ$ which means $LM \parallel ID$ and $LN \parallel IE$. This gives us that that $LM=LN$. Now $\angle LMN = \angle IDE = \angle LBC$ which means that there is a spiral similarity at $L$ taking $MN \longrightarrow BC \implies P \in \odot(LBC) \equiv \odot(ABC) \ \blacksquare$