Dear Mathlinkers, acording to the statements, D is the point of contact of the incircle of ABC and BC... Sincerely Jean-Louis

Note that through length chasing $D$ is indeed the point where the incircle is tanget to $BC$ and that $AD$ is the radical axis of $w_1$, $w_1$. Hence, it suffices to prove that $AE=AG$ which is true through length chasing.

Good problem. Blastzit wrote: In $\Delta ABC$, let $D$ be a point on side $BC$. Suppose the incircle $\omega_1$ of $\Delta ABD$ touches sides $AB$ and $AD$ at $E,F$ respectively, and the incircle $\omega_2$ of $\Delta ACD$ touches sides $AD$ and $AC$ at $F,G$ respectively. Suppose the segment $EG$ intersects $\omega_1$ and $\omega_2$ again at $P$ and $Q$ respectively. Show that line $AD$, tangent of $\omega_1$ at $P$ and tangent of $\omega_2$ at $Q$ are concurrent. Let $I,I_1,I_2$ denote the incenters of$\triangle ABC, \triangle ABD$ and $\triangle ACD$ respectively. Also, let the tangents at $P$ and $Q$ meet in $T$. We prove the subsequent claims: $ $ Claim 1:$D$ is the in-touch point of side $BC$ of $\triangle ABC$. Proof. Let $AE = AF = AG =x$. Then we have that $CY = CG = b-x$ and $BX = BE = c-x$, where $X,Y$ respectively are the foot of perpendiculars from $I_1,I_2$ on $BC$. Then, we easily get that $D$ is the mid-point of $XY \implies BD = s-b$ as desired. Now, let $EI_1 \cap GI_2 = L, EG \cap I_1I_2=H$. Then, clearly $AELG$ is cyclic. So, $$\angle LAE = \angle LGE = \angle I_2GQ = \frac{1}{2} \cdot \angle A \iff A-I-L$$Consequently, applying Desargue's theorem on $\triangle EBI_1$ and $\triangle GCI_2$, we have that $H \in \overline{BC}$. Claim 2: $T$ lies on line $AD$. $ $ Proof. Observe that $H$ is the ex-similicenter of $\omega_1$ and $\omega_2$ which means that $H : \triangle EOP \mapsto \triangle QRG \iff \angle EOP = \angle QRG$ where $O,R$ are the in-touch points of $BD$ and $CD$. So, we have $$ \angle TPQ = \angle EFP = \angle EOP = \angle QRG = \angle QFG = \angle TQP \iff TP = TQ$$ Thus, $$ TI_1^2 - I_1F^2 = \text{Pow}(T, \omega_1) = TP^2 = TQ^2 = \text{Pow}(T, \omega_2) = TI_2^2 - I_2F^2 \iff TF \perp I_1I_2 \iff T \in \overline{AF} \blacksquare$$ Hamel wrote: Note that through length chasing $D$ is indeed the point where the incircle is tanget to $BC$ and that $AD$ is the radical axis of $w_1$, $w_1$. Hence, it suffices to prove that $AE=AG$ which is true through length chasing. May I know why is that condition sufficient, rest aside the fact that $AE$ is (obviously) equal to $AG$

e_plus_pi wrote: May I know is that condition sufficient, rest aside the fact that $AE$ is (obviously) equal to $AG$ Yes. It suffices to prove that the intersection of the tangets $X$ is on $AD$ (hence that point must have equal powers to both circles. Thus the two angles forming the triangle $PXQ$ must be equal , which is equal to proving that angle $AEG$ is equal to $AGE$ )

Invert at $F$ with an arbitrary radius. The problem turns to as shown in the picture (where the primes denote inverted points) We only have to prove $P'Q' \perp A'F$, which is trivial because $P',Q'$ are symmetric about perpendicular bisector of $E'G'$, which is parallel to $A'F$

Attachments:

First let's construct a circle $\omega$ tangent to $\omega_1$ at $P$ passing through $Q$, such that the tangent to $\omega_2$ intersects $\omega$ at $Q_1$. Notice that if we can prove $Q=Q_1$, we'll be done by Radical Axis, as then all circles will be mutually tangent, thus all of the tangents must concur. Now let the $2$ tangents intersect at $X$. As $\angle XPQ = \angle PQ_1Q=\angle P_1X$, to prove $Q=Q_1$, we need to prove $\angle XPQ = \angle XQP$. Let $D_1$ be the intersection of $\omega_1$ with $BC$ and $D_2$ of $\omega_2$ with $BC$, and $\angle A= \alpha$, $\angle B = \beta$, $\angle C= \gamma$. Notice $\angle AEP = 90 - \frac{\alpha}2$ and $\angle BED_1 = 90 - \frac{\beta}2 => \angle D_1EP = 90 -\frac{\gamma}2$. Thus $\angle D_1PX=\angle D_1EP = 90 -\frac{\gamma}2$, and $\angle EPD_1 = \angle ED_1B = 90 -\frac{\beta}2$, thus $XPQ=90 -\frac{\alpha}2$. Similarly, we get $\angle XQD_2 = 90 -\frac{\beta}2$ and $\angle D_2QG=90 -\frac{\gamma}2=>\angle XQP = 90 - \frac{\alpha}2$. Thus $\angle PQX=\angle PQ_1X=>Q=Q_1$, thus we're done.

jayme wrote: Dear Mathlinkers, acording to the statements, D is the point of contact of the incircle of ABC and BC... Sincerely Jean-Louis Certainly this is a true fact but it is irrelevant regarding the solution of this problem. You just show that the tangents at $P$ and $Q$ intersect at a point $Y$ and $YP=YQ$ (by angle chasing, considering $\angle{AEF} = \angle{AFE} $ ). Consequently $Y$ is in the radical axis of the two circles, which is the line $AD$.