Let $T \in BC$ be such that $\angle TAD=90^{\circ}$ and let $O_1, O_2$ denote the circumcenters of $\triangle TAB, \triangle TAC$. Then $$\angle O_1AB=\angle O_1BA=90^{\circ}-\angle ATB=\angle ADB, $$so $O_1A, O_1B$ touch $(ABD)$. Similarly, $O_2A, O_2C$ touch $(ACD)$. So, both circles pass through $T$.

cute Art problem $\iff$ impossible for my geogebra addicted mind

VicKmath7 wrote: I undertood that the point of intersection coincides with this point T that you mentioned. But how could one realize that TÂD = 90°?

It's hard to write the sol than to make it Let BC hit circle (O2) at H So we have to prove H also lies on (O1) Firstly angle AHD is 1/2 angle AO2C and angle ADH = angle ACO2 Now drop perp from O2 on AC let that point be E. Note that triangle AHD is similar to O2EC By angle chasing we get angle AO1B = angle AO2C and similarly when we drop a perp from O1 on AB call it F hence O1FB is similar to AHD hence as angle AHD = 1/2 Angle AO1B we are done. On the contrary one can also use phantom points like let BC hit (O1) at H' and follows... I don't know why but I can't latex it

@kayqm One can just guess by the diagram, or do simple angle chasing with $\angle AXC$, where $X$ is the intersection of the circle with center $O_{1}$ and line $BC$.

Really cute but easy opening problem! Funnily i guessed that this intersection point would be the point mentioned in above posts, but I ended up finding a solution without proving it. The crux of the solution is the following key claim. Claim : Triangle $\triangle ABC$ and $\triangle AO_1O_2$ are similar. Proof : First we notice that, \[\measuredangle O_2AO_1 = \measuredangle DAO_1 + \measuredangle O_2AD = \measuredangle CAB\]Now we denote by $X_1$ and $X_2$ the circumcenters of $\triangle ABD$ and $\triangle ADC$. Note that, \[\measuredangle BX_1A = 2\measuredangle BDA = \measuredangle CX_2A\]which implies that $\triangle ABX_1 \sim \triangle ACX_2$ (since they are both clearly isosceles). Now, since it is clear that $AO_1BX_1$ and $AO_2CX_2$ are cyclic, \[\measuredangle AO_1X = \measuredangle ABX_1 = \measuredangle ACX_2 = \measuredangle AO_2X\]which then implies that $\triangle AO_1X_1 \sim \triangle AO_2X_2$ (since these triangles are both clearly right). Now, \[\frac{AO_1}{AO_2}=\frac{AX_1}{AX_2}= \frac{AB}{AC}\]which implies the claim. Now with this claim in hand, let $P$ denote the second intersection of the circles $\omega_1$ centered at $O_1$ passing through $A$ , and $\omega_2$ centered at $O_2$ passing through $A$. Note that due to symmetry $\overline{OO_1}$ is the internal $\angle PO_2A$-bisector. Thus, \[\measuredangle ACP = \measuredangle AO_2O_1 = \measuredangle ACB\]implying that points $B$ , $C$ and $P$ are collinear, as we set out to show.

Let $\omega_{1}$ be the circle with center $O_{1}$ and $\omega_{2}$ be defined similarly. Letting $E$ be the second intersection of $\omega_{2}$ with line $BC$, and $F$ the opposite point of $A$ in $\omega_{1}$ and $H$ similarly in $\omega_{2}$. We need only to prove $$90^\circ - \angle FAB = \angle AFB = \angle AEB$$We can see, by a simple angle seek, $$\angle AEB = \angle AEC = \angle AHC = 90^\circ - \angle HAC$$Therefore, the problem becomes to prove that $$\angle FAB = \angle HAC$$which is trivial since both are equal to $\angle ADB$ by tangency.

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