Let $BC=a, CA=b, AB=c$ 1. $\Rightarrow$ $\angle BAX=\angle MAC\Rightarrow \dfrac{a+c-b}{a+b-c}=\dfrac{c^2}{b^2}$ $\Rightarrow (b-c)(a(b+c)-b^2-c^2)=0$ Since $b\neq c$, $a(b+c)=b^2+c^2$ $\Rightarrow b(a-b)+c(a-c)=0$ $\Rightarrow b(a+c-b)=c(b+c-a)$, which implies if $D_1\in BC$ satisfies $ZD_1//AC$, then $ZD_1=ZA$. Similarly, if $D_2\in BC$ satisfies $YD_2//AB$, then $YD_2=YA$. Besides, $BD_1+CD_2=\dfrac{a(a+c-b)}{2c}+\dfrac{a(a+b-c)}{2b}=\dfrac{a(ab+ac-b^2-c^2+2bc)}{2bc}=a$, which implies $D_1=D_2$. Since $AYD_1Z$ is a rhombus, we have $D=D_1$ and $YZ$ is the perpendicular bisector of $AD$. 2. $\Leftarrow$ Since $AYDZ$ is a rhombus, we have $AC//ZD$ and $AB//YD$. $\Rightarrow BD+CD=a+\dfrac{a}{2bc}(ab+ac-b^2-c^2)=a\Rightarrow ab+ac=b^2+c^2$ $\Rightarrow ab+ac+bc=b^2+bc+c^2$ $\Rightarrow (b-c)(ab+ac-bc)=b^3-c^3$ $\Rightarrow ab^2+b^2c-b^3=ac^2+bc^2-c^3$ $\Rightarrow \dfrac{c^2}{b^2}=\dfrac{a+c-b}{a+b-c}=\dfrac{BD}{DC}$, which implies $\angle BAX=\angle MAX$. So we're done.

Too easy if you are familiar with projective geometry Ida: Let F be the intersection of the tangents to (ABC) by B and C, let G be the intersection of $ZY$ with $AD$ and $E$ be the intersection of the tangent of A with BC, as $CZ, BY ,AX$ coincides in the gergonne stitch, so being $E'=ZY \cap BC$ we have $(E', X; B,C)=-1$ but as AX is simmedian, being $R=AX\cap (ABC)$ we have that $ ABCR$ is harmonic, so projecting by $ A \Rightarrow (E, X;B, C)=-1\Rightarrow E=E'$, by angle chasing we get $\angle EAD= \angle EAB+\angle BAD=\angle DCA+\angle DAC=\angle EDA$ which implies $EAD$ isosceles and as $EG$ is height we have that $G$ is the midpoint of AD. Volta: If G is the midpoint of $AD$, being $E=ZY\cap BC$ we have that $(E,X; B,C)=-1$ and tracing $EA$ we have that $\angle EAD=\angle EAB+\angle BAD=\angle CAD+\angle EAB=\angle EDA =\angle CAD+\angle DCA\Rightarrow \angle EAB=\angle ACB \Rightarrow EA$ is tangent to $(ABC)$, hence projecting by $A$ we have that if $H=AX\cap (ABC)$, then $ABHC$ is harmonic $\Rightarrow AX$ is symmedian.

Let $YZ$ meet $AD$ at $N$. Let $K$ be midpoint of $AD$. $AK = \frac{AD}{2} = \frac{bc}{b+c}.\cos{\frac{A}{2}}$ and $AN = (p-a).\cos{\frac{A}{2}}$ so we need to prove $p-a = \frac{bc}{b+c}$ or $b+c-a = \frac{2bc}{b+c}$ or $b^2 + c^2 = a(b+c)$ Claim $: b^2 + c^2 = a(b+c)$ Proof $:$ Note that $\frac{CM}{BM} = \frac{\sin{MAC}}{\sin{MAB}}.\frac{b}{c} \implies \frac{\sin{MAC}}{\sin{MAB}} = \frac{c}{b}$. Note that $\frac{\sin{XAB}}{\sin{XAC}} = \frac{\sin{MAC}}{\sin{MAB}}$ so $\frac{BX}{CX} = \frac{\sin{XAB}}{\sin{XAC}}.\frac{c}{b} = \frac{c^2}{b^2}$ so $\frac{a+c-b}{a+b-c} = \frac{c^2}{b^2} \implies (b-c)(a(b+c)-b^2-c_2) = 0$ and since $ABC$ is scalen then $a(b+c)-b^2-c^2 = 0 \implies b^2 + c^2 = a(b+c)$.

Solved with Om245. An exercise in harmonics! Just to be complete we prove both directions but they are exactly the same. Forward implication ( $\angle BAX=\angle MAC$) : It is not hard to see that this angle condition simply says that $X$ lies on the $A-$symmedian. Let the second intersection of the $A-$symmedian and $(ABC)$ be $R$. Further, let the intersection of lines $\overline{YZ}$ and $\overline{BC}$ be $T$. Then, we know that $(AR;BC)=-1$. But, we also know that, \[-1=(TX;BC)\overset{A}{=}(A'R;BC)\]where $A'$ is the second intersection of line $\overline{AT}$ with $(ABC)$. Combining this with the previous observation it follows that $AT$ must be a tangent to $(ABC)$. Now, notice that this implies, \[\measuredangle DAT = \measuredangle DAB + \measuredangle BAT = \measuredangle CAD + \measuredangle BCA = \measuredangle BDA = \measuredangle TDA\]from which it follows that $\triangle TAD$ is isosceles. But, it is clear that $YZ$ is the $T$-altitude of this triangle since it is well known that $YZ$ is perpendicular to the $A$-angle bisector. Since the altitude between the two equal sides of an isosceles triangle is also a perpendicular bisector, it follows that $\overline{YZ}$ bisects $AD$ as desired. Reverse Implication ($YZ$ passes through the midpoint of $AD$. ) Proof : It is clear that now, in addition to being the $T$-altitude, $YZ$ is also the perpendicular bisector of $AD$. Thus, $T$ is equidistant to points $A$ and $D$ which implies, \[\measuredangle BAT = \measuredangle DAT + \measuredangle BAD = \measuredangle TDA + \measuredangle BAD = \measuredangle BDA + \measuredangle DAC = \measuredangle DCA = \measuredangle BCA\]from which it follows that $AT$ is a tangent to $(ABC)$. Now, since we know that, \[-1=(TX;BC) \overset{A}{=}(AR;BC)\]from which it is clear that $X$ lies on the $A$-symmedian as desired.