Anyone? :0

I'll post a proof for a), b) is not difficult. Define lines parallel to $AB,BC,CA$ through $X.$ One of these lines $l$ must be closer to $XY$ than $XZ$ is, otherwise all lines are contained in a $90^\circ$ angle (made by the internal and external angle bisectors of $\angle XYZ$ and containing $XZ$), implying the triangle is not acute. If $XY \ge XZ,$ then this implies that $X_iY_i > X_iZ_i$ for some particular $i$ since $XY$ makes a smaller acute angle with the side of $ABC$ parallel to $l$ than $XZ$ does. $\blacksquare$

a) Claim 1: any line can be uniquely placed to pass through a vertex of $ABC$ and intersect the opposite segment in the interior. Proof: Construct $\ell$ parallel to $BC$ passing through $A$. Using parallel lines, the sectors around $A$ partition the space of arguments for the line. Claim 2: Say that a line $\ell_1$ is closer to a side $PQ$ than line $\ell_2$ if the angle between $\ell_1,PQ$ is smaller than the angle between $\ell_2, PQ$. Then, in an acute triangle $\triangle ABC$, it is impossible for $XZ$ to be closer to each side than $XY$. Proof: Using Claim 1, WLOG, we may place $X=B$ and $Y\in AC$. Then, reflect $Y$ across $BA,BC$ to $Y_C, Y_A$. AFTSOC that $XZ$ is closer to $BC$, this implies that that $Z \in \angle YBY_C$ (or the other side), and assuming $XZ$ is closer to $BA$ implies that $Z \ in \angle Y_C BY$ (or the other side). The only way these sectors intersect is if $180 < \angle Y_C B Y + \angle YBY_C = 2\angle ABC$, which is impossible because $\angle B$ is acute, and so it is impossible for $XZ$ to be closer to both $BC$ and $BA$. This finishes (a) because if $XZ< XY$, then there exists a side for which $XZ$ is further (angle-wise). Combining these, this gives $X_i Z_i < X_i Y_i$ where $i$ corresponds to the side where $XZ$ is further angle-wise, a contradiction. $\blacksquare$. b) Consider a triangle where $\angle A = 180 - \varepsilon$. Then, if $XZ$ is parallel to $BC$ and $XY$ is perpendicular to $BC$, one can have $XZ<XY$ while still having $X_iZ_i >> X_i Y_i$ for all $i$. $\blacksquare$.

I used trigonometry for this problem. a) Case 1: $X$ lies on the left of $Y$ and $Z$. First, let $XX_1$ intersect $YY_3$ at $P$, $XX_1$ intersect $ZZ_3$ at $Q$, and $YY_1$ intersect $ZZ_3$ at R. As trapeziums $X_3XPY_3$ and $X_3XQZ_3$ are similar, $XP/XQ$ = $X_3Y_3/X_3Z_3$. In particular, $XP < XQ$. Similiarly, $QZ > QR = PY$. Next, as $Z_3QX_1B$ is cyclic, we have \begin{align*} \angle X_1BZ_3 &= 180 - \angle X_1BZ_3 \\ &= 180 - \angle XQZ \\ &= 180 - \angle XPY \end{align*}. Since $\angle X_1BZ_3$ is acute, $\angle XPY$ is obtuse. So, \[ \cos(\angle XPY) < 0 \]. Now, by cosine rule, \begin{align*} (XY)^2 &= (XP)^2 + (PY)^2 - 2(XP)(PY)\cos(\angle XPY) \\ &< (XQ)^2 + (QZ)^2 - 2(XQ)(QZ)\cos(\angle XQZ) \\ &= (XZ)^2 \end{align*}. So $XY < XZ$. $\Box$ Case 2: $X$ lies between $Y$ and $Z$ This case can be handled similarly as case 1 except with a different sets of points. $\Box$ QED b) Let $\angle CBA$ be the obtuse angle, then it is possible to find a counter example by letting $XY$ perpendicular to $AB$. QED