http://www.artofproblemsolving.com/Forum/viewtopic.php?t=307

hophinhan wrote: (Romania, 1997) Let ABC be a triangle, let D be a point on side BC, and let (O) be the circumcircle of triangle ABC. Show that the circles tangent to (O),AD,BD and to (O),AD,DC are tangent to each other if and only if \hat{BAD} = \hat{CAD}. Let I be the incenter of triangle ABC. +If AD is the bisector of $ \angle BAC$, let $ AD\cap (O) = \{E\}$ $ \Rightarrow E$ is the midpoint of arc BC Then $ M,N,E$ are collinear. We have $ EM.EN = EC^2 = EI^2$ Construct a tangency EK of $ (O_2) (K\in AE)$ then $ EK^2 = EI^2$ so $ K\equiv I$ So $ (O_1)$ and $ (O_2)$ are tangent to each other at I. +If $ (O_1)$ and $ (O_2)$ are tangent to each other. By Sawayama-Thébault's theorem and widen Lyness's theorem, it's easy to prove that they are tangent at I. Therefore $ \angle BAD = \angle CAD$

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Dear Mathlinkers, you can also see another way at http://perso.orange.fr/jl.ayme vol. 4 A new mixtilinear incircle adventure 3 p. 9-10 with a reference. Sincerely Jean-Louis

Please correct me if I say something wrong or stupid. But I think that the problem would be "better" if we are able to prove this problem: PROBLEM-Let $ ABC$ be a triangle. Let $ D$ be the point on $ BC$. Let $ (O)$ be the circumcircle of $ \triangle ABC$. Show that the $ (O_1)$, $ (O_2)$ tangent to each other at $ T$, where $ (O_1)$ is the circle that tangents to $ (O),AD,BD$, $ (O_2)$ is the circle that tangents to$ (O),AD,CD$, if and only if $ T$ is the incircle of $ \triangle ABC$. Indeed, with problem that hophinhanLHP has been proposed, we have a generalized problem for it, which is: LEMMA-Given $ (O)$ and $ A,B$ be $ 2$ points that lie on $ (O)$. Let $ (O_1),(O_2)$ be circles that lie on the same side wrt $ AB$ such that they both internallu tangent to $ (O)$ respectively at $ X_1,X_2$. Let $ \{C,D\}\equiv (O_1) \cap (O_2)$. Let $ K$ be the point that lie on the different side with $ (O_1),(O_2)$ wrt $ AB$. Prove that $ K$ is the midpoint of arc $ AB$ if and only if $ K\in CD$. Proof: Let $ r,r_1,r_2$ be the radii of $ (O),(O_1),(O_2)$, respectively.$ K_1,K_2$ respectively be the tangency point of $ (O_1),(O_2)$ with $ AB$. Consider the homothety through point $ X_1$, ratio $ k_1 = \frac {r}{r_1}$. We have $ \mathcal {H}(X_1,k_1): (O_1)\mapsto (O)$, $ K_1\mapsto K'\Longrightarrow K'\in (O)$. Because $ O_1K_1\bot AB\Longrightarrow OK'\bot AB$, which implies that $ K'$ is the midpoint of arc $ AB$. With the same argument for $ (O_2)$, we conclude that $ X_1K_1,X_2K_2$ intersect at the point $ K\in (O)$ such that $ K$ is mipoint of arc $ AB$. We will prove that this will equivalent to $ K\in CD$. Indeed, when $ K$ is the midpoint of arc $ AB$, it is easy to prove that $ KA^2 = \overline {KK_1}\cdot \overline {KX_1}$, $ KB^2 = \overline {KK_2}\cdot \overline {KX_2}$. But $ KA = KB\Leftrightarrow \overline {KK_1}\cdot \overline {KX_1} = \overline {KK_2}\cdot \overline {KX_2}\Leftrightarrow \mathcal {P}_{K/(O_1)} = \mathcal {P}_{K/(O_2)}\Leftrightarrow K\in$ the radical axes of $ (O_1),(O_2)\Leftrightarrow K\in CD$. The result is lead as follow. Back to the beginning problem. Since $ A,D,T$ are collinear which is equivalent to $ \overline {ADT}$ is the radical axes of $ (O_1),(O_2)$. Let $ K\equiv AD\cap (O)$ ($ K$ is different from $ A$) $ \Leftrightarrow K$ is the midpoint of arc $ BC\Leftrightarrow AT$ is the bisector of $ \angle BAC$. It is easy to note that $ KB = KT = KC$. Therefore $ \angle KBT = \angle KTB$. But $ \angle KBD\equiv \angle KBC = \angle KAC = \angle BAD$, which implies that $ \angle TBA = \angle TBC\Leftrightarrow BT$ is the bisector of $ \angle ABC$. Thus $ T$ is the incircle of $ \triangle ABC$.

We'll use this little lemma : "If $(O;R);(O';R') [R>R']$ is internally tagent at point $K$ .Let $AB$ be a chord of $(O)$ and tagent to $(O')$ at $E$ then $KE$ passes through the midpoint of arc $AB$ "(Easy to prove , right ?) Let $J$ be midpoint of arc$BC$ not having $A$ . Let $(O_1) ;(O_2)$ be the circles tagent to $(O);DA;DB$ and $(O);DA;DC$ respectively. Call $E,F$ are tagency points of $(O_1);(O_2)$ with $(O)$ respectively . $(O_1);(O_2) $ tagent to $BC$ at $P,Q$ respectively According to the lemma , we have : $J,E,P$ are collinear ; $J,Q,F$ are collinear. Furthermore , we have :$ \overline{JE}.\overline{JP}=JB^2 ; \overline{JQ}.\overline{JF}=JC^2$ $\Rightarrow \overline{JE}.\overline{JP}=\overline{JQ}.\overline{JF}$ $\Rightarrow J$ lies on the racial axis of $(O_1);(O_2)$. Due to that we get : $ (O_1);(O_2)$ are tagent to each other $ \Leftrightarrow AD $ is their radical axis $\Leftrightarrow J \in AD$ $\Leftrightarrow \angle BAD =\angle CAD$ Done

Lemma-Let $ABC$ be a triangle.$I$ its incenter and $D$ a point on $BC$.consider the circle that is tangent to the circumcircle of $ABC$ but is also tangent to $DC$,$DA$ at $E,F$ respectively. Then $E,F$ and $I$ are collinear. now in problem,if both circles are tangent together but $\angle BAD$ and $\angle CAD$ are different then clearly $I$ is not lies on $AD$.but the tangent points of the two circles collinear with $I$.which cant be possible(a tangent point is common from them).hence $BD$ is angle bisector.

Let $\omega_1$ be the circle which is tangent to $BC, AD$ and $\mathcal{O}$. Define $\omega_2$ similarly. Let $\omega_1$ touches $BC, AD$ at $P, Q$ and $\omega_2$ touches $BC, AD$ at $R, S$ respectively. Let $I$ be the incenter of $\triangle ABC$. By the Curvilinear Incircle lemma we know $P, I, Q$ are collinear. Similarly $R, S, I$ are collinear. Suppose $D$ is the foot of $A-$angle bisector. Then $I=Q$ and similarly $I=S$. So $I$ will be the touch point of $\omega_1$ and $\omega_2$ on $AD$. Therefore $\omega_1, \omega_2$ will be tangent. Now suppose $\omega_1, \omega_2$ are tangent on $AD$ which is $Q=S$, call it $I'$. Since by curvillinear we have $I$ lies on $PI'$ and $I$ lies on $RI'$ therefore $I=I'$ and the tangency point of these two circle is actually the incenter. Since the tangency point lies on $AD$ therefore $I$ lies on $AD$. As desired.