9)

Решение:

В треугольнике ABC:

\( \angle ACB = 180^{\circ} - 15^{\circ} - 45^{\circ} = 120^{\circ} \).

CM — медиана, значит \( AM = MC = MB \).

Треугольник AMC равнобедренный, \( \angle MAC = \angle MCA = 45^{\circ} \).

\( \angle BCM = \angle ACB - \angle ACM = 120^{\circ} - 45^{\circ} = 75^{\circ} \).

В треугольнике BMC:

\( \angle CBM = 15^{\circ} \).

\( \angle CMB = 180^{\circ} - 15^{\circ} - 75^{\circ} = 90^{\circ} \).

Значит, \( \triangle BMC \) — прямоугольный.

CM — медиана к гипотенузе AB.

По теореме Пифагора в \( \triangle ABC \): \( AB^2 = AC^2 + BC^2 \).

В \( \triangle AMC \), по теореме синусов:

\( \frac{AC}{\sin 45^{\circ}} = \frac{AM}{\sin 45^{\circ}} = \frac{MC}{\sin 15^{\circ}} \Rightarrow AC = AM = \frac{MC}{\sin 15^{\circ}} \cdot \sin 45^{\circ} \).

\( AC = MC \cdot \frac{\sin 45^{\circ}}{\sin 15^{\circ}} = MC \cdot \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{6}-\sqrt{2}}{4}} = MC \cdot \frac{2\sqrt{2}}{\sqrt{6}-\sqrt{2}} \cdot \frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}} = MC \cdot \frac{2\sqrt{12} + 4}{6-2} = MC \cdot \frac{4\sqrt{3} + 4}{4} = MC(\sqrt{3}+1) \).

В \( \triangle BMC \) прямоугольном:

\( MC = BC \sin 15^{\circ} \Rightarrow BC = \frac{MC}{\sin 15^{\circ}} = MC \cdot \frac{4}{\sqrt{6}-\sqrt{2}} \cdot \frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}+\sqrt{2}} = MC \cdot \frac{4(\sqrt{6}+\sqrt{2})}{4} = MC(\sqrt{6}+\sqrt{2}) \).

\( x = MB = MC \).

\( BM = MC \).

\( \angle BAC = 45^{\circ} \).

\( AB = 2 MC \).

\( AC = MC(\sqrt{3}+1) \).

\( BC = MC(\sqrt{6}+\sqrt{2}) \).

\( x = BM = MC \).

Ответ: \( MC \)