Let the sides of triangle $ABC$ be $a$, $b$, $c$ and the sides of triangle $XYZ$ be $x$, $y$, $z$. Let the medians of triangle $ABC$ be $g_a$, $g_b$, $g_c$ to sides $a$, $b$, $c$ respectively, and the medians of triangle $XYZ$ be $g_x$, $g_y$, $g_z$ to sides $x$, $y$, $z$ respectively. We are given that the ratios of the medians are proportional: $g_a : g_b : g_c = g_x : g_y : g_z$ This implies that there exists a constant $k$ such that: $g_x = k g_a$, $g_y = k g_b$, $g_z = k g_c$ We know from Apollonius' Theorem that the length of a median in a triangle is given by: $g_a^2 = \frac{2b^2 + 2c^2 - a^2}{4}$ $g_b^2 = \frac{2a^2 + 2c^2 - b^2}{4}$ $g_c^2 = \frac{2a^2 + 2b^2 - c^2}{4}$ Similarly, for triangle $XYZ$: $g_x^2 = \frac{2y^2 + 2z^2 - x^2}{4}$ $g_y^2 = \frac{2x^2 + 2z^2 - y^2}{4}$ $g_z^2 = \frac{2x^2 + 2y^2 - z^2}{4}$ Substituting $g_x = k g_a$, $g_y = k g_b$, and $g_z = k g_c$, we have: $k^2 g_a^2 = \frac{2y^2 + 2z^2 - x^2}{4}$ $k^2 g_b^2 = \frac{2x^2 + 2z^2 - y^2}{4}$ $k^2 g_c^2 = \frac{2x^2 + 2y^2 - z^2}{4}$ Substituting the expressions for $g_a^2, g_b^2, g_c^2$, we get: $k^2 (2b^2 + 2c^2 - a^2) = 2y^2 + 2z^2 - x^2 \hspace{1cm} (1)$ $k^2 (2a^2 + 2c^2 - b^2) = 2x^2 + 2z^2 - y^2 \hspace{1cm} (2)$ $k^2 (2a^2 + 2b^2 - c^2) = 2x^2 + 2y^2 - z^2 \hspace{1cm} (3)$ Adding equations (1), (2), and (3), we get: $k^2 (3a^2 + 3b^2 + 3c^2) = 3x^2 + 3y^2 + 3z^2$ $k^2 (a^2 + b^2 + c^2) = x^2 + y^2 + z^2 \hspace{1cm} (4)$ Subtracting equation (1) from (2): $k^2(2a^2 - 2b^2) = 2x^2 - 2y^2$ $k^2(a^2 - b^2) = x^2 - y^2 \hspace{1cm} (5)$ Subtracting equation (2) from (3): $k^2(2b^2 - 2c^2) = 2y^2 - 2z^2$ $k^2(b^2 - c^2) = y^2 - z^2 \hspace{1cm} (6)$ From (5), we have $x^2 - y^2 = k^2(a^2 - b^2)$. From (6), we have $y^2 - z^2 = k^2(b^2 - c^2)$. Adding these two equations, we get: $x^2 - z^2 = k^2(a^2 - c^2) \hspace{1cm} (7)$ From (5), we have $\frac{a^2 - b^2}{x^2 - y^2} = \frac{1}{k^2}$. From (6), we have $\frac{b^2 - c^2}{y^2 - z^2} = \frac{1}{k^2}$. From (7), we have $\frac{a^2 - c^2}{x^2 - z^2} = \frac{1}{k^2}$. This implies: $\frac{a^2 - b^2}{x^2 - y^2} = \frac{b^2 - c^2}{y^2 - z^2} = \frac{a^2 - c^2}{x^2 - z^2}$ Using the given ratios: $\frac{g_a}{g_x} = \frac{g_b}{g_y} = \frac{g_c}{g_z} = \frac{1}{k}$ Squaring, we get: $\frac{g_a^2}{g_x^2} = \frac{g_b^2}{g_y^2} = \frac{g_c^2}{g_z^2} = \frac{1}{k^2}$ This leads to: $\frac{2b^2+2c^2-a^2}{2y^2+2z^2-x^2} = \frac{2a^2+2c^2-b^2}{2x^2+2z^2-y^2} = \frac{2a^2+2b^2-c^2}{2x^2+2y^2-z^2} = \frac{1}{k^2}$ From equation (4), $x^2 + y^2 + z^2 = k^2 (a^2 + b^2 + c^2)$, we get $\frac{a^2 + b^2 + c^2}{x^2 + y^2 + z^2} = \frac{1}{k^2}$. We have: $\frac{a^2}{x^2} = \frac{b^2}{y^2} = \frac{c^2}{z^2} = \frac{a^2+b^2+c^2}{x^2+y^2+z^2}$ Hence, the triangles $ABC$ and $XYZ$ are similar.