Inversion at $D$ kills part(a).

General problem for part a) https://artofproblemsolving.com/community/c374081h1619335 And general problem for part b) https://artofproblemsolving.com/community/c374081h1619336

General problem for part a: Let $B,C$ be two point lying outside $(J)$ such that $BC$ is tangent to $(J)$ at $D$. Draw two tangents $CE,BF$ to $(J)$. An arbitrary circle through $D,J$ meets $DE,DF$ at $H,K$. Then $(BDH), (CDK), (J)$ are coaxial. Proof. Draw $CS\perp JH$, $BR\perp JK$, $CS$ meets $BR$ at $X$. We have $\angle BXC=180^\circ-\angle RJS=\angle KDH=180^\circ-\angle BJC.$ Then $KBJC$ is cyclic. Using Simson line theorem we have $R,D,S$ are collinear. Let $U$ be the second intersection of $RS$ with $(J)$. $XU$ intersects $(J)$ at $V.$ We will show that $(BDH), (CDK)$ pass through $V$. Since $H$ lies on polar of $C$ wrt $(J)$ then $C$ lies on polar of $H$, we get $CS $ is the polar of $H$. Then $H$ and $S$ are inverse wrt $(J).$ From this, we get $\overline{JH}\cdot\overline{JS}=JD^2=JU^2$, which follows that $U,D,H,J$ are concyclic. Hence $\angle UHD=\angle UJD=2\angle DEU$, then $HU=HE$. We obtain $UE\perp JS$. Let $N$ be the midpoint of $UE$. Let $W$ be a point on $AV$ such that $HW\parallel XC\parallel UE$. We have $\angle WHD=\angle UED=\angle UVD$, then $WDHV$ is cyclic. Draw $JT\perp DU.$ Applying Menelaus theorem for triangle $UNS$ with 3 collinear points $D,H,E$ we conclude that $\dfrac{\overline{HN}}{\overline{HS}}=\dfrac{\overline{DU}}{2\overline{DS}}=\dfrac{\overline{DT}}{\overline{DS}}.$ Since $\triangle ABJ\sim\triangle SDJ$ then $\dfrac{\overline{DT}}{\overline{DS}}=\dfrac{\overline{BR}}{\overline{BA}}$. Therefore $\dfrac{\overline{HN}}{\overline{HS}}=\dfrac{\overline{BR}}{\overline{BA}}$. Using Thales theorem we have $BW\parallel RU$. Then $\angle WBD=\angle BDR=\angle UVD$, which follows that $BWDV$ is cyclic. Therefore $B,W,D,H,V$ are concyclic. Do the same with another circle we are done. For part b we don't need $90^\circ$, the result is still true when $AP, AQ$ are isogonal wrt $\angle BAC.$ Proof. We know that $SI$ passes through the tangency $ W$ of $A$-mixtilinear incircle of triangle $ABC$ with $(O)$. On the other side, $AW$ and $AD$ are isogonal wrt $\angle BAC$. So we only need to prove that $AT$ and $AD$ are isogonal wrt $\angle BAC.$ Lemma. Let $ABC$ be a triangle. $P,Q$ be two points such that $AP,AQ$ are isogonal wrt $\angle BAC$. $BP$ meets $CQ$ at $X$, $BQ$ meets $CP$ at $Y.$ Then $AX,AY$ are also isogonal wrt $\angle BAC.$ This lemma is well-known so I leave for the readers. Back to problem. Let $I_b, I_c$ be the excenters wrt $B,C$ of triangle $ABC$. $CI_c$ meets $JG$ at $U$. We have $CJ,CU$ are bisector of $\angle NCG$ then $(JUNG)=-1$. On the other side, $(BCDG)=-1$ then $BJ, CU,DN$ concur at $I_c.$ Similarly we have $M,D,I_b$ are collinear. $PN$ meets $QM$ at $R$. Since $AI_c$ and $AI_b$, $AP$ and $AQ$ are isogonal wrt $\angle BAC$ then $A(PI_cJN)=A(QI_bJM)$ or $N(API_cJ)=M(AQI_bJ)$, which follows that $A,R,D$ are colllinear. Applying the lemma above for triangle $AMN$ we have $AP,AQ$ are isogonal wrt $\angle MAN$, $NP$ meets $MQ$ at $R$, $MP$ meets $NQ$ at $T$ then $AT, AR$ are isogonal wrt $\angle BAC$. We are done.

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Here is my solution for this problem Solution a) Let $A'$, $B'$, $C'$, $L'$, $E'$, $F'$, $J'$, $K'$, $H'$ be images of $A$, $B$, $C$, $L$, $E$, $F$, $J$, $K$, $H$ through the inversion $I_D^k$ Then: these sets of points $(A', B', F', D)$, $(A', C', E', D)$ lie on a circle; $(B', F', J')$, $(C', E', J')$, $(L', H', J', K')$ are collinear Let $U'$ $\equiv$ $B'K'$ $\cap$ $H'C'$ Applying Pappus theorem for 2 sets of 3 collinear points $(B', D, C')$ and $(H', J', K')$, we have: $F'$ $\equiv$ $B'J'$ $\cap$ $DH'$, $E'$ $\equiv$ $C'J'$ $\cap$ $DK'$, $F'$ $\equiv$ $B'K'$ $\cap$ $H'C'$ are collinear or $B'K'$, $H'C'$, $E'F'$ concurrent So: through the inversion, $I_D^k$, $(BDK)$, $(HDC)$, $(J)$ have 2 common points b) It's easy to prove that: $AD$, $BE$, $CF$ concurrent Then: $(BCDG) = J(BCDG) = J(PQDG) = - 1$ Let $V$ be midpoint of $PQ$ We have: $\dfrac{d(A; JP)}{d(A; JQ)} = \dfrac{AB . \sin \widehat{ABP}}{AC . \sin \widehat{ACQ}} = \dfrac{\sin C . \cos \dfrac{B}{2}}{\sin B . \cos \dfrac{C}{2}} = \dfrac{\sin \dfrac{C}{2}}{\sin \dfrac{B}{2}} = \dfrac{\cos \widehat{ACQ}}{\cos \widehat{ABP}} = \dfrac{AF . \cos \widehat{ACQ}}{AE . \cos \widehat{ABP}} = \dfrac{JP}{JQ}$ So: $JA$ is $J$ - symmedian of $\triangle$ $JPQ$ But: $JA$, $JD$ are isogonal conjugate in $\widehat{PJQ}$ then: $JD$ is $J$ - median of $\triangle JPQ$ or $J$, $D$, $V$ are collinear Hence: $J(PQDG) = J(PQVG) = - 1$ or $\overline{M, J, N, G}$ $\parallel$ $PQ$ Let $R$ $\in$ $AT$ which satisfies $MR$ $\parallel$ $AP$ We have: $\dfrac{\overline{TM}}{\overline{TP}} = \dfrac{\overline{TN}}{\overline{TQ}} = \dfrac{\overline{TR}}{\overline{TA}}$ Then: $NR$ $\parallel$ $AQ$ So: $(RN; RM) \equiv (AQ; AP) \equiv (AC; AB) \equiv (AN; AM)$ (mod $\pi$) or $A$, $N$, $R$, $M$ lie on a circle But: $AM$ $\perp$ $RM$ and $AD$ $\perp$ $MN$ then $AR$, $AD$ are isogonal conjugate in $\widehat{BAC}$ Let $W$ $\equiv$ $SI$ $\cap$ $(O)$ $(W \not \equiv S)$ so: $W$ is tangency point of $A$ - mixtilinear incircle with $(O)$ Hence: $AW$, $AD$ are isogonal conjugate in $\widehat{BAC}$ Therefore: $A$, $W$, $R$, $T$ are collinear or $AT$ intersects $SI$ at a point lies on $(O)$

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A bit messy, but still enjoyable. We relabel points a little bit. Replace $J, K, H,L$ with $I_A, E_1, F_1,M$ respectively. (a) Let $O_B, O_C$ be the centers of $\odot(BDE_1)$ and $\odot(CDF_1)$. We want to show that $O_B, I_A, O_C$ are colinear. Thus, we take the homothety $\mathcal{H}(I_A, 2)$ and let $O_B\mapsto B_1$ and $O_C\mapsto C_1$. It's easy to see that $BB_1$ and $CC_1$ are perpendicular to $BC$ (which follows from that $O_B$ lies on the perpendicular bisector of $BD$). $MB_1\perp DE\parallel I_AC$ and similarly $MC_1\parallel I_AB$ (which follows from that $O_B$ lies on the line passing through the midpoint of $I_AM$ and perpendicular to $DE$). Surprisingly, we are one step away. Apply Pappus' Theorem on $\overline{BMC}$ and $\overline{{\infty}_{I_AC}{\infty}_{\perp BC}{\infty}_{I_AB}}$ gives $I_A = \overline{B{\infty}_{I_AB}}\cap \overline{C{\infty}_{I_AC}}$, $B_1 = \overline{M{\infty}_{I_AB}}\cap\overline{B{\infty}_{\perp BC}}$ and $C_1 = \overline{M{\infty}_{I_AC}}\cap\overline{C{\infty}_{\perp BC}}$ are colinear. (b) By Mixtilinear's properties, it suffices to show that $\angle BAT = \angle CAT$. First, we note that $G$ is the pole of $AD$ w.rt. $\odot(I_A)$. Thus $MN\perp AD$. Now we claim that Claim: $PQ\parallel MN$. Proof: Extend $AD$ to meet $\odot(ABC)$ at $X$. We will prove that $\triangle PAQ\stackrel{-}{\sim}\triangle CXB$ through length bashing. Note that $\triangle BI_AF\sim\triangle BPA$ thus $$AP = \frac{I_AF\cdot AB}{BF} = \frac{cr_a}{s-c} \implies \frac{AP}{AQ} = \frac{c(s-b)}{b(s-c)}.$$But by ratio lemma, we get $\tfrac{BX}{XC} = \tfrac{b(s-c)}{c(s-b)}$ hence the similarity follows. Now the result follows from $$\measuredangle APQ = \measuredangle BCX = \measuredangle BAD = 90^{\circ} + \angle PAD \blacksquare$$ From here, the problem is destroyed by applying DDIT twice. Let $U = PN\cap QM$ and apply DDIT: on the quadrilateral $PNMQ$, we get an involution swapping $(AM, AP)$, $(AN, AQ)$, $(AU, A\infty_{\perp AD})$. Keeping in mind that rotating by $90^{\circ}$ is an involution, we get $A, U, D$ are colinear. on the quadrilateral $PMQN$, we get an involution swapping $(AP, AQ)$, $(AM, AN)$, $(AT, AU)$. Thus this involution must be isogonality. Combining with the above application, we are done.

Part (a): Invert at $D$. Same as all above. Part(b): Claim: $AD \perp PQ$ . Let $AD \cap PQ = D'$. Since $\frac{\sin \angle BAD}{ \sin \angle CAD} = \frac{BD}{DC} \cdot \frac{\sin \angle B}{ \sin \angle C}= \frac{\tan \angle B/2}{\tan \angle C/2} \cdot \frac{\sin \angle B}{ \sin \angle C}$ and $\frac{\cos \angle PAD'}{\cos \angle QAD'} = \frac{PA}{QA} = \frac{\sin \angle C}{\tan \angle B / 2} \bigg/ \frac{\sin \angle B}{\tan \angle C / 2}$ implies the result. Let $AD$ intersect $PQ$ and $MN$ at $X,Y$ respectively. Since $AD$ is the polar of $G,$ $AD \perp MN$. From the claim, $AD \perp PQ$. Thus, $PQ \parallel MN$. Not only that, $\triangle PXA \sim \triangle AYM$ and $\triangle QXA \sim \triangle AYN$, so \[\frac{PX}{QX} = \frac{PX}{XA}\cdot \frac{XA}{QX} = \frac{AY}{MY}\cdot \frac{NY}{AY} = \frac{NY}{MY}.\]Therefore, $PN,QM,AD$ are concurrent at a point $U$. That is, $U$ lies on $AD.$ Let $V$ be the isogonal conjugate of $U$ with respect to $\triangle AMN$. By Jacobi's theorem on $\triangle AMN$, $PM,QN,AV$ are concurrent. Hence, $A,V,T$ are collinear. Since $U,V$ are isogonal conjugate, $AT$ and $AU$ are isogonal lines of $\angle BAC$. Obviously, $AT$ intersect $\odot(ABC)$ at $A-$mixtillinear incircle touch-point, which is also $SI\cap AT$.