Let $(I_1)\cap AB=Z,W$ and $(I_2)\cap AC=X,Y$ where $Z,E,W$ and $X,F,Y$ collinear in this order. We've $\angle{NEM}=\angle{NFM}\implies NEFM$ concyclic. This gives $AX\times AY=AE\times AM=AF\times AN=AZ\times AW$. So, $WXZY$ is cyclic. Since $\angle{DEZ}=180^{\circ}-\angle{DEB}=\angle{DZN}\implies DZ^2=DE\times DN$ and $\angle{DWN}=\angle{DFC}=\angle{DEW}\implies DW^2=DE\times DN$. We get that $DW=DZ$. Hence $D$ lie on perpendicular bisector of $WZ$. Similarly, $D$ lie on perpendicular bisector of $XY$. This gives us $D$ must be the centre of $(WXZY)$. Let $M_1$ and $M_2$ denote the midpoint of minor arc $FN$ and $EM$ of $(I_1)$ and $(I_2)$, respectively. It's well-known that $D,K,M_1$ and $D,H,M_2$ are collinear. It's also easy to see that $\angle{M_1DN}+\angle{M_2DM}=\frac{\angle{EDF}}{2}+\frac{\angle{EDF}}{2}=\angle{EDF}$, so $D,M_1,M_2$ collinear. Hence, $D,M_1,M_2,K,H$ are all collinear. We've It's well-known that $DK\times DM_2=DX^2=DY^2$ (since $D$ is the midpoint of arc $XY$ of $(I_2)$). Similarly, $DM_1\times DH=DZ^2$. Comparing two equations gives us $DM_1\times DH =DK\times DM_2$. Also, note that $\triangle{DJ_1K}\sim \triangle{DI_1M_1}$ and $\triangle{DJ_2H}\sim \triangle{DI_2M_2}$. We get $\frac{DJ_1}{DI_1}=\frac{DK}{DM_1}=\frac{DH}{DM_2}=\frac{DJ_2}{DI_2}\implies J_1J_2\parallel I_1I_2$. Hence, $DQ$ which is radical axis of $(J_1)$ and $(J_2)$, perpendicular to $J_1J_2$, is the same line with $DP$ which is radical axis of $(I_1)$ and $(I_2)$, perpendicular to $I_1I_2$. This finishes part a. Let $R=HK\cap EF$. Note that by Miquel's point of quadrilateral, we get that $ERGH$ concyclic. Since $A$ lie on radical axis of $(I_1)$ and $(I_2)$, we get that $D,A,P,Q$ are all collinear. Let $EF$ intersect the tangent line at $D$ of $(DQG)$ at $T$. We've $\angle{GRD}=180^{\circ}-\angle{GHK}=180^{\circ}-\angle{GEA}=180^{\circ}-\angle{GLA}=\angle{GLD}$. This means $R\in (DLG)$. And since $\angle{GRT}=\angle{GHE}=\angle{GQL}=\angle{GDT}$, $GRDT$ is cyclic. In other words, $T$ lies on $(DLG)$. This finishes part b.

gausskarl wrote: We have a scalene acute triangle $ABC$ (triangle with no two equal sides) and a point $D$ on side $BC$. Pick a point $E$ on side $AB$ and a point $F$ on side $AC$ such that $\angle DEB=\angle DFC$. Lines $DF,\, DE$ intersect $AB,\, AC$ at points $M,\, N$, respectively. Denote $(I_1),\, (I_2)$ by the circumcircles of triangles $DEM,\, DFN$ in that order. The circle $(J_1)$ touches $(I_1)$ internally at $D$ and touches $AB$ at $K$, circle $(J_2)$ touches $(I_2)$ internally at $D$ and touches $AC$ at $H$. $P$ is the intersection of $(I_1),\, (I_2)$ different from $D$. $Q$ is the intersection of $(J_1),\, (J_2)$ different from $D$. a. Prove that all points $D,\, P,\, Q$ lie on the same line. b. The circumcircles of triangles $AEF,\, AHK$ intersect at $A,\, G$. $(AEF)$ also cut $AQ$ at $A,\, L$. Prove that the tangent at $D$ of $(DQG)$ cuts $EF$ at a point on $(DLG)$. $a)$ first remark that $EFMN$ is cyclic so $ A$ is on the radical axis of $(I_1),(I_2)$ ;let the tangents of $(I_1),(I_2)$ cut $AB,AC$ at $C' ,B'$ respectively.it s easy to show that $AB'DC'$ is parallelogram thus $AK=AH$ then $A$ is also on the radical axis of $(J_1),(J_2)$ hence $P,Q,D,A$ are collinear. $b)$ $ \angle C'KD=\angle KDC', \angle DHB' =\angle B'DH ,\angle C'DB'= \angle C'AB'$ thus $K,H,D $ are collinear ,consider $T$ the intersection of $EF $ and $KH$ complete quadrilateral property leads $TGFH $ is cyclic so $\angle GTD =\angle GFA =\angle GLA $ thus $T \in (DLG)$ . $ \angle STD =\angle SLA =\angle GLA -\angle GLS=\angle GFA -\angle GDS= \angle GFA- \angle GQD =\angle GFA-\angle GQA=\angle GFA-\angle GHA=\angle FGH $ since $ Q \in (AKH) [\because \angle AQH=\angle DQH=\angle DHA=\angle AKH]$ so$\angle STD=\angle FGH = \angle FTH $ which means that $T,S,F$ are collinear hence the result follows.

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Here is my solution for this problem Solution a) Let $I$, $J$ be midpoint of $\stackrel\frown{ME}$, $\stackrel\frown{NF}$ then it's well - known that: $D$, $H$, $J$ are collinear and $D$, $K$, $I$ are collinear So: $\dfrac{J_2D}{I_2D} = \dfrac{DH}{DJ}$ and $\dfrac{J_1D}{I_1D} = \dfrac{DK}{ID}$ Hence: $\dfrac{DH}{HJ} = \dfrac{DN}{NJ} . \dfrac{DF}{FJ} = \dfrac{DN . DF}{NJ^2}$ Similarly: $\dfrac{DK}{KI} = \dfrac{ED . DM}{MI^2}$ Then: $\dfrac{DH}{HJ} : \dfrac{DK}{KI} = \dfrac{DN . DF}{NJ^2} : \dfrac{ED . DM}{MI^2} = \dfrac{DN . DF}{ED . DM} . \dfrac{MI^2}{NJ^2}$ Since: $\angle{DEB} = \angle{DFC}$ so: $M$, $N$, $E$, $F$ lie on a circle Then: $\triangle DFN$ $\sim$ $\triangle DEM$ or $\dfrac{DF}{ED} = \dfrac{DN}{DM} = \dfrac{FN}{EM} = \dfrac{NJ}{MI}$ So: $\dfrac{DH}{HJ} = \dfrac{DK}{KI}$ or $\dfrac{DH}{DJ} = \dfrac{DK}{ID}$ Hence: $\dfrac{J_2D}{I_2D} = \dfrac{J_1D}{I_1D}$ or $J_1J_2$ $\parallel$ $I_1I_2$ But: $J_1J_2$ $\perp$ $QD$ then: $QD$ $\perp$ $I_1I_2$ Combine with: $DP$ $\perp$ $I_1I_2$ so: $D$, $P$, $Q$ are collinear b) Let point $S$ on $EF$ which satisfies $\dfrac{\overline{SE}}{\overline{SF}} = \dfrac{\overline{DK}}{\overline{DH}}$ We have: $\angle{GKH} = \angle{GAH} = \angle{GAF} = \angle{GEF}$ and $\angle{GHK} = \angle{GAK} = \angle{GFE}$ So: $\triangle GKH$ $\sim$ $\triangle GEF$ Then: $\dfrac{GK}{KH} = \dfrac{GE}{EF}$ or $\dfrac{GK}{GE} = \dfrac{KH}{EF}$ Since: $\dfrac{SE}{SF} = \dfrac{DK}{DH}$, we have: $\dfrac{SE}{DK} = \dfrac{SF}{DH} = \dfrac{SE - SF}{DK - DH} = \dfrac{EF}{HK}$ Hence: $\dfrac{EG}{KG} = \dfrac{EF}{KH} = \dfrac{SE}{DK}$ But: $\angle{GKH} = \angle{GEF}$ so: $\triangle DGK$ $\sim$ $\triangle SGE$ Then: $\angle{DGK} = \angle{SGE}$ and $\dfrac{KG}{DG} = \dfrac{EG}{SG}$ Hence: $\angle{SGD} = \angle{EGK}$ or $\triangle GKE$ $\sim$ $\triangle GDS$ So: $\angle{GKE} = \angle{GDS} = \angle{GQD}$ or $DS$ tangents $(DQG)$ at $D$ Since: $\triangle GKE$ $\sim$ $\triangle GDS$ then: $\angle{DSG} = \angle{KEG}$ But: $\angle{KEG} = \angle{AEG} = \angle{ALG}$ so: $\angle{DSG} = \angle{ALG}$ or $D$, $L$, $G$, $S$ lie on a circle Hence: tangent at $D$ of $(DQG)$ intersects $EF$ at a point $S$ lies on $(DLG)$