Denote by $O_1$ the center of $\omega_1$ and by $O_2$ the center of $\omega_2$. From Thebault's Theorem we obtain that $O_1,I,O_2$ are collinear. Consequently, $I$ is the midpoint of $O_1O_2$. Finally,(since $O_1DMO_2$ is cyclic) we get that $ID=IM$ ($M$ -the midpoint of $BC$), thus $2a=b+c$.

Can you detail more your soltution? It seems to be a great solution. My solutions uses Cayley th twice and a lot of calculations... Your solution seems to avoid the calculations. The thebault theorem I found sais: On the sides of parallelogram ABCD erect squares -- all either on the outside or the inside of the parallelogram. Their centers then form another square. I don't see where is the link with this problem.

Actually, Thebault's Theorem states that points $O_1,I,O_2$ are collinear. A lemma used in its proof is the following fact: If $P$ is the midpoint of $M_1M_2$ (these are the tangency points of the two circles with $BC$) then $IP||AD$. Thus, in this case, we get that $I$ is the midpoint of $O_1O_2$. Finally, it is enough to notice that: \[ \angle{O_1DO_2}=\angle{O_1MO_2}=90^{\cdot} \] and the conclusion follows.

Lemma 1. let $w_1$ be a circle which is interior tangent in the point $A$ to the circle $w$. For a point $M\in w_1$, $M\ne A$ denote the intersections $X,Y$ between the circle $w$ and the tangent-line in the point $M$ to the circle $w_1$. Then the ray $[AM$ is the bisector of the angle $\widehat {XAY}$. Lemma 2. For the incenter $I$ and the centroid $G$ of the triangle $ABC$ we have $\blacktriangleright IG\parallel BC\Longleftrightarrow b+c=2a\ .$ $\blacktriangleright IG\perp BC\Longleftrightarrow b=c\ \ \vee\ \ b+c=3a\ .$ Indication for proof the proposed problem. The circles $w_1,w_2$ are interior tangent to the circle $w$. You can apply the above remarkable lemma for each of the circles $w_1,w_2$ and for each of the lines $BC,AD,l$.

Every one has used this theoerm < Thebault's Theorem> can some give link or its proof Abdurashid

Sailor wrote: If $P$ is the midpoint of $M_{1}M_{2}$ (these are the tangency points of the two circles with $BC$), then $IP \| AD$. Could you (or anyone else) give a proof of this fact? The rest is quite clear, but I can't figure this part.

I didn't know the statement Sailor said, but I was pretty sure that the general statement of Thebault theorem says that: $ O_{1}, I, O_{2}$ are collinear and $ \frac {O_{1}I}{O_{2}I} = \tan^{2}{\frac {\alpha}{2}}$, where $ \angle{ADB} = \alpha$, where $ AD$ is the "Thebault cevian". In this case, $ \alpha = 90$, so the conclusion follows. Here you can see, the statement of this thing, followed by another nice ones. http://links.jstor.org/sici?sici=0002-9890%28200204%29109%3A4%3C362%3ATAOTGP%3E2.0.CO%3B2-A&size=SMALL&origin=JSTOR-reducePage PS: sorry for the long link, i googled it.

Here is a full solution to this problem using Sawayama's Theorem and some length chasing. [asy][asy] size(250); defaultpen(linewidth(0.7)+fontsize(11pt)); pair A=dir(120),B=dir(210),C=dir(330),D=foot(A,B,C),I=incenter(A,B,C),Ap=(3*A+2*D)/5, Dp=1.4*D-0.4*A; path alt = Ap--Dp; real r1=0.222, r2=0.708; pair O1=(D.x-r1,D.y+r1), O2=(D.x+r2,D.y+r2); path c1=circle(O1,r1),c2=circle(O2,r2); draw(A--B--C--A--D^^unitcircle^^c1^^c2); pair M=(foot(O1,B,C)+foot(O2,B,C))/2, Mp=foot(M,O1,O2); path ellp = reflect(M,Mp)*alt; pair F=foot(O1,reflect(M,Mp)*A,reflect(M,Mp)*D),P1=intersectionpoint(O1--F,c1); path ell=shift(P1-F)*ellp; pair P=intersectionpoint(ell,B--C); draw(incircle(A,B,C)^^foot(O1,B,C)--I^^foot(O2,B,C)--foot(O2,A,D),linetype("4 4")); draw(rightanglemark(A,D,C,2)); dot(I^^foot(I,B,C)^^O1^^O2); draw(ell); label("$A$",A,dir(origin--A)); label("$B$",B,dir(origin--B)); label("$C$",C,dir(origin--C)); label("$D$",D,S); label("$P$",P,S); label("$F_1$",foot(O1,B,C),S); label("$E_1$",foot(O1,A,D),dir(160)); label("$F_2$",foot(O2,B,C),S); label("$E_2$",foot(O2,A,D),dir(20)); label("$I$",I,N); label("$T$",foot(I,B,C),S); label("$O_1$",O1,dir(95)); label("$O_2$",O2,N); [/asy][/asy] WLOG let $AB<AC$. We first define several auxiliary points. Let $O_1$ and $O_2$ denote the centers of $\omega_1$ and $\omega_2$ respectively. Let the points of tangency of $\omega_1$ with $AD$ and $BD$ are $E_1$ and $F_1$ respectively; define $E_2$ and $F_2$ in a similar manner. Denote by $I$ the incenter of $\triangle ABC$, and let $T$ be the point where its incircle is tangent to $BC$. Furthermore, let $P=\ell\cap BC$. Finally, let $P_1$ and $P_2$ be the points of tangency of $\ell$ with $\omega_1$ and $\omega_2$ respectively. By Sawayama's Theorem, $E_1F_1$ passes through the incenter $I$ of $\triangle ABC$. Since $DE_1=DF_1$, $\angle IF_1F_2=45^\circ$. Similarly, $\angle IF_2F_1=45^\circ$. Therefore $\triangle IF_1F_2$ is an isosceles right triangle, implying that $T$ is the midpoint of $\overline{F_1F_2}$. Now we introduce a lemma: LEMMA: $P$ is the reflection of $D$ over $T$. Proof. First examine $\omega_1$. Since $O_1F_1\perp BC$ and $O_1P_1\perp\ell$, quadrilateral $O_1F_1PP_1$ is cyclic, and combining this with $O_1F_1=O_1P_1$ gives that $PO_1$ bisects $\angle F_1PP_1$. Similarly, $PO_2$ bisects $\angle F_2PP_2$. As a result, $O_1P\perp O_2P$, which implies $\angle O_1PF_1$ and $\angle O_2PF_2$ are complementary $\implies \triangle O_1PF_1\sim\triangle PO_2F_2$. Now let $r_1$ and $r_2$ be the radii of $\omega_1$ and $\omega_2$ respectively. Remark that $PF_1+PF_2=F_1F_2=r_1+r_2$ from the fact that $AD\perp BC$ and that $PF_1\cdot PF_2=r_1r_2$ from the above similarity. Thus, we may conclude that $\{PF_1,PF_2\}=\{r_1,r_2\}$. However, since $AC>AB$, $r_2>r_1$, so the line $\ell$ will slope in such a way that $PF_1>PF_2$. Thus $PF_1=r_2$ and $PF_2=r_1$. Finally, note that $DF_1=r_1$ and $DF_2=r_2$; this reversal of lengths implies that $P$ and $D$ are reflections about the midpoint of $\overline{F_1F_2}$, which is $T$. $\blacksquare$ This allows us to completely get rid of $\omega_1$ and $\omega_2$ and turn this problem entirely into a length-chasing one. Assume general triangle notation, so that $a=BC$, etc. We now prove the "only if" condition of the problem statement and then show that all our steps are reversible, thus proving the "if" condition as well. ONLY IF: Suppose that $P$ is indeed the midpoint of $\overline{BC}$. Then $BP=CP=\tfrac a2$. Furthermore, it is well-known that $BT=s-b$ and $CT=s-c$. Therefore \[BD=2\cdot BT-BP=2(s-b)-\dfrac a2=a+b+c-2b-\dfrac a2=\dfrac a2+c-b.\] Similarly, $CD=\tfrac a2+b-c$. Now from the Pythagorean Theorem $AB^2-BD^2=AD^2=AC^2-CD^2$. Plugging in everything we know gives \begin{align*}c^2-\left(\dfrac a2+c-b\right)^2&=b^2-\left(\dfrac a2-b+c\right)^2\\4c^2-(a+2c-2b)^2&=4b^2-(a+2b-2c)^2\\4(c^2-b^2)&=(a+2c-2b)^2-(a+2b-2c)^2\\&=[a+2c-2b-(a+2b-2c)][a+2c-2b+(a+2b-2c)]\\&=8a(c-b)\\\implies b+c&=2a\end{align*} as desired. IF: Now that all the algebra above has been hacked out, the "if" clause is easy. Start with $b+c=2a$, and multiply both sides by $4(c-b)$. Reversing all our steps above leads to the crucial equality \[c^2-\left(\dfrac a2+c-b\right)^2=b^2-\left(\dfrac a2-b+c\right)^2.\] Remark that the converse of our use of the Pythagorean Theorem is true as well; in other words, if $AB^2-BD^2=AC^2-CD^2$, then $AD\perp BC$. In this case, let $D'$ be the point for which $BD'=\tfrac a2+c-b$ and $CD'=\tfrac a2-b+c$. By the perpendicularity criterion, $AD'\perp BC$. Furthermore, since \[\left(\dfrac a2+c-b\right)+\left(\dfrac a2-b+c\right)=2\left(\dfrac a2\right)=a,\] points $B$, $D'$, and $C$ are collinear. As a result, $D'\equiv D$. Finally, we can reverse our original algebra using our Lemma to get $BP=CP=\tfrac a2$, proving this direction as well. Now we have proven both directions, so we are done. $\blacksquare$

My solution: Let $ I $ be the Incenter of $ \triangle ABC $ and $ T= \odot (I) \cap BC $ . Let $ I_1 $ be the center of $ \omega_1 $ and $ T_1=\omega_1 \cap BC, S_1=\omega_1 \cap AD $ . Let $ I_2 $ be the center of $ \omega_2 $ and $ T_2=\omega_2 \cap BC, S_2=\omega_2 \cap AD $ . Let $ O $ be the circumcenter of $ \triangle ABC $ and $ M=AI \cap \odot (ABC), K= \ell \cap BC $ . From here we get $ 2BC=AB+AC \Longleftrightarrow AI=IM $ . ... $ (1) $ From Sawayama's Theorem we get $ I \in S_1T_1, I \in S_2T_2 $ , so $ \angle IT_1T_2=\angle IT_2T_1=45^{\circ} \Longrightarrow T $ is the midpoint of $ T_1T_2 $ , hence combine with $ T_1D=KT_2 $ (well-known) we get $ T $ is the midpoint of $ DK $ . ... $ (2) $ From $ (1), (2) $ we get $ K $ is the midpoint of $ BC \Longleftrightarrow 2BC=AB+AC $ . Q.E.D

sailor wrote: Finally, it is enough to notice that: \[ \angle{O_1DO_2}=\angle{O_1MO_2}=90^{\cdot} \] Why is $\angle{O_1MO_2}=90$? Thanks.

Does anyone know why? Thanks in advance.

Quote: Why is $\angle{O_1MO_2}=90$? Thanks. $P$ is of course a projection of $I$ on $AB$. Now since $I$ is a midpoint of $O_1O_2$, we get that $P$ is a midpoint of $M_1M_2$. Besides due to fact that $OI \perp AI$, we have $PD=PM$, so if $IM_1=IM_2$, then $ID=IM$ q.e.d.

Dear Mathlinkers, a link Ayme J.-L., Sawayama and Thébault's theorem, Forum Geometricorum (2003) ; http://forumgeom.fau.edu/ Sincerely Jean-Louis

Pinionrzek wrote: Quote: Besides due to fact that $OI \perp AI$. Can you explain this in more detail? Because that would mean that ,if $X$ is the midpoint of arc $BC$ not containing $A$,then $AI$=$IX$,and since $IX=BX = \frac{BC}{2\cos \frac{A}{2}}$ and $AI$=$\frac{\frac{b+c-a}{2}}{\cos \frac{A}{2}}$ we would directly get the conclusion.

Here's the easy explanation for why $\angle O_1MO_2 = 90$. We are assuming that the internal tangent other than $AD$ intersects $BC$ at $M$. Let this other internal tangent touch $\omega_1$ at $X_1$ and touch $\omega_2$ at $X_2$. Then, $O_1M$ bisects $\angle X_1MB$ and $O_2M$ bisects $\angle X_2MC$, so then $\angle O_1MO_2$ must be $90$.