I only had solved the first part of the problem: $\angle BAC=\angle CBD=\angle BCD =\angle BAG+\angle GAC$ Let $M$, $X$ and $Y$ the midpoints of $BC$, $BE$ and $CF$, respectively. $BC$ is tangent to the circumcircle of $ABG$ at $B$, and by power of a point: $$MB^2=MG(MA)=MC^2$$ $BC$ is tangent to the circumcircle $ACG$ at $C$. The reflection $G^\prime$ of $G$ relative to $M$ lies on the circumcircle of $ABC$ (because $MB^2=MG(MA)=G^\prime M(MA)=BM(BC)$). Aditionally, $BGCG\prime$ is a parallelogram ($BC$ and $GG\prime$ intersect at $M$, the midpoint of these segments). All the properties implies: $$\angle CAG=\angle BAC - \angle BAG^\prime=\angle BCD-BCG^\prime=\angle GEC$$$$\angle BAG=\angle BAC - \angle CAG^\prime=\angle CBD-CBG^\prime=\angle GFB$$ $$\angle AEC=\angle ACB, \angle ACE=\angle ABC, \angle CAE=\angle BAC$$ $$\angle AFB=\angle ABC, \angle ABF=\angle ACB, \angle BAF=\angle BAC$$ So, we need to prove that $\angle MAN=180^\circ -\angle BDC=2 \angle BAC$ . If we prove that $\frac{FM}{MB}=\frac{CN}{NE}$, $AFM\sim ACN$ and $AMB\sim ANE$, in other words: $\angle MAB+\angle CAN+\angle ABC=2\angle BAC$. Finally, by Menelaus's theorem: $$\frac{FM}{MB}\cdot\frac{BX}{XG}\cdot\frac{GY}{YF}=1\implies \frac{FM}{MB}=\frac{XG}{BX}\cdot\frac{YF}{GY}$$$$\frac{CN}{NE}\cdot\frac{EX}{XG}\cdot\frac{GY}{YC}=1\implies \frac{CN}{NE}=\frac{XG}{BX}\cdot\frac{YF}{GY}$$ I need help to solve the second part of the problem. I found that }$H$ is the intersection of the reflections of $GA$, $GB$ and $GC$ to the bisectors of the angles $BAC$, $ABC$ and $ACB$, respectively; $ABKC$ is a parallelogram.

a. Define $T$, $X$ and $Y$ as the midpoint of $BC$, $BE$ and $CF$ respectively. Define $\Omega_B$ as a circle through $A$ which tangent to $BC$ at $B$. Define $\Omega_C$ similarly. It's well known that $\Omega_B$ and $\Omega_C$ pass through $G$. Notice that \[\angle AGE=\angle BAG+\angle GBA=\angle GBM+\angle GBA=\angle ABC=\angle ECA\]Hence, we must have $E\in \Omega_C$. Similarly, $F\in \Omega_C$. Therefore, by spiral similarity, we have $\triangle ABE\thicksim \triangle AFC$. As $X$ is midpoint of $BE$ and $Y$ is midpoint of $CF$, then we have $\triangle ABX\thicksim \triangle AFY$. As $\triangle ABX\thicksim \triangle AFY$, then we have $AYGX$ and $AXBM$ are both cyclic (by spiral similarity). Analogously, we can check that $AYCN$ is cyclic. Therefore, \[\measuredangle AMD=\measuredangle AXG=\measuredangle AYG=\measuredangle AND\]So, $AMDN$ is cyclic. b. It is easy to check that $H$ is the isogonal conjugate of $G$ wrt $\triangle ABC$. ($G$ is the $A$-humpty point while $H$ is the $A$-dumpty point). We will prove a generalization of this problem. $\textbf{Generalization:}$ Let $ABC$ be a triangle and $P$ be a point on the plane. Define $Q$ as the isogonal conjugate of $P$ wrt $\triangle ABC $. Let $X_A$ as the second intersection of $AP$ and $\odot (BPC)$. We define $X_B$ and $X_C$ similarly. Let $Y_A$ as the intersection of the perpendicular bisector of $X_BX_C$ and $BC$. Define $Y_B$ and $Y_C$ similarly. Prove that $Y_A,Y_B,Y_C$ are collinear. $\textbf{Proof:}$ Define $Z_A, Z_B, Z_C$ as the circumcenter of $\triangle QX_BX_C, \triangle QX_CX_A$ and $\triangle QX_AX_B$ respectively. By Desargues' it suffices to show that $AZ_A$, $BZ_B$ and $CZ_C$ are concurrent. Define $U$ as the circumcenter of $\triangle X_AX_BX_C$. $\textit{Claim:}$ Isogonal conjugate of $U$ wrt $\triangle ABC$ lies on $AZ_A$ $\textit{Proof:}$ Do inversion with center $A$ and radius $\sqrt{AB\times AC}$ continued by reflection across internal angle bisector of $\angle BAC$. It's easy to check that this process send $X_B$ to $X_C$ and $Q$ to $X_A$ (Both converse are also true). Suppose this process send $Z_A$ to $W$. As $Z_A$ is the circumcenter of $QX_BX_C$, then $W$ lies on the line through $A$ and the circumcenter of $X_AX_BX_C$ which is $U$. So, $U$ lies on $AW$. Furthermore, $AW$ is isogonal to $AZ_A$ wrt $\angle ABC$. So, the isogonal conjugate of $U$ wrt $\triangle ABC$ lies on $AZ_A$. (Claim proven) By the claim, then we have $AZ_A, BZ_B, CZ_C$ will concur at the isogonal conjugate of $U$ which then finish the problem.

We have: $\textbf{Lemma:}$ Given points $X,X',Y,Y'$ such that $X$ and $X'$, $Y$ and $Y'$ are images of one another by $\sqrt{bc}$ inversion (inversion and reflection about angle bisector), we have that the centers of $(XYY')$ and $(X'YY')$ are isogonal wrt. $\angle BAC$. $\textbf{Proof:}$ Obviously the two circles are images of one another by $\sqrt{bc}$ inversion. Let $\omega$ be the image of $(XYY')$ by inversion around $A$ with radius $\sqrt{AB\cdot AC}$. We obviously have that $(ABC), \omega$, and $(XYY')$ are coaxial, so their centers are collinear, and since $\omega$ reflects about the bisector of $\angle BAC$ to $(X'YY')$, we are done. Using some of what @above said for isogonal conjugates and inversion in the configuration, this immediately implies part b) by Jacobi.