Question:

Step 1: Analyze the geometric properties due to the angle bisector and median. Let $O$ be the intersection of the angle bisector $AK$ and the median $BM$. Given that $AK$ is the angle bisector of $\angle A$ and $AK \perp BM$. In $\triangle ABM$, $AO$ is the angle bisector of $\angle BAM$ and $AO$ is an altitude to $BM$. This implies that $\triangle ABM$ is an isosceles triangle with $AB = AM$. Also, $AO$ is the median to $BM$, meaning $O$ is the midpoint of $BM$. Since $BM$ is a median of $\triangle ABC$, $M$ is the midpoint of $AC$. Therefore, $AM = MC$. From $AB = AM$ and $AM = MC$, we have $AB = AM = MC$. Let $AB = c$. Then $AM = c$ and $MC = c$. Thus, the side $AC = AM + MC = c + c = 2c$. So, in $\triangle ABC$, we have $AB=c$, $AC=2c$, and $\angle ABC = 120^\circ$. Step 2: Apply the Law of Sines to find $\sin C$. In $\triangle ABC$, using the Law of Sines: $$ \frac{AB}{\sin C} = \frac{AC}{\sin B} $$ Substitute the known values: $AB=c$, $AC=2c$, and $\angle B = 120^\circ$. $$ \frac{c}{\sin C} = \frac{2c}{\sin 120^\circ} $$ $$ \sin C = \frac{c \sin 120^\circ}{2c} = \frac{\sin 120^\circ}{2} $$ Since $\sin 120^\circ = \sin(180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$: $$ \sin C = \frac{\sqrt{3}/2}{2} = \frac{\sqrt{3}}{4} $$ Step 3: Calculate $\cos C$. Since $\angle B = 120^\circ$, angles $A$ and $C$ must be acute (as $A+C = 180^\circ - 120^\circ = 60^\circ$). Therefore, $\cos C$ must be positive. Using the trigonometric identity $\sin^2 C + \cos^2 C = 1$: $$ \cos C = \sqrt{1 - \sin^2 C} = \sqrt{1 - \left(\frac{\sqrt{3}}{4}\right)^2} $$ $$ \cos C = \sqrt{1 - \frac{3}{16}} = \sqrt{\frac{13}{16}} = \frac{\sqrt{13}}{4} $$ Step 4: Determine $\angle A$ and calculate $\cos A$. The sum of angles in a triangle is $180^\circ$: $$ A + B + C = 180^\circ $$ $$ A = 180^\circ - B - C = 180^\circ - 120^\circ - C = 60^\circ - C $$ Now, calculate $\cos A$ using the angle subtraction formula $\cos(X - Y) = \cos X \cos Y + \sin X \sin Y$: $$ \cos A = \cos(60^\circ - C) = \cos 60^\circ \cos C + \sin 60^\circ \sin C $$ Substitute the values of $\sin C$ and $\cos C$: $$ \cos A = \left(\frac{1}{2}\right)\left(\frac{\sqrt{13}}{4}\right) + \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{3}}{4}\right) $$ $$ \cos A = \frac{\sqrt{13}}{8} + \frac{3}{8} = \frac{3+\sqrt{13}}{8} $$ Step 5: Apply the Law of Cosines to find side $BC$ (let $a=BC$) in terms of $AB$ (let $c=AB$) and confirm Option 1. In $\triangle ABC$, using the Law of Cosines for side $a=BC$: $$ a^2 = AB^2 + AC^2 - 2(AB)(AC)\cos A $$ Substitute $AB=c$, $AC=2c$, and $\cos A = \frac{3+\sqrt{13}}{8}$: $$ a^2 = c^2 + (2c)^2 - 2(c)(2c)\left(\frac{3+\sqrt{13}}{8}\right) $$ $$ a^2 = c^2 + 4c^2 - 4c^2\left(\frac{3+\sqrt{13}}{8}\right) $$ $$ a^2 = 5c^2 - c^2\left(\frac{3+\sqrt{13}}{2}\right) $$ Factor out $c^2$: $$ a^2 = c^2 \left(5 - \frac{3+\sqrt{13}}{2}\right) $$ $$ a^2 = c^2 \left(\frac{10 - 3 - \sqrt{13}}{2}\right) $$ $$ a^2 = c^2 \left(\frac{7 - \sqrt{13}}{2}\right) $$ To find the ratio $\frac{BC}{AB} = \frac{a}{c}$: $$ \left(\frac{a}{c}\right)^2 = \frac{7 - \sqrt{13}}{2} $$ We can check if $\left(\frac{\sqrt{13}-1}{2}\right)^2$ matches this value: $$ \left(\frac{\sqrt{13}-1}{2}\right)^2 = \frac{(\sqrt{13})^2 - 2\sqrt{13}(1) + 1^2}{4} = \frac{13 - 2\sqrt{13} + 1}{4} = \frac{14 - 2\sqrt{13}}{4} = \frac{7 - \sqrt{13}}{2} $$ Since the values match, $$ \frac{a}{c} = \frac{\sqrt{13}-1}{2} $$ Thus, the value of ratio $\frac{BC}{AB}$ is equal to $\frac{\sqrt{13}-1}{2}$. Option 1 is correct. Step 6: Calculate the ratio of the radius of the circumscribing circle $R$ to the side length $AB$ and confirm Option 2. Using the Extended Law of Sines, we know that for any triangle $\frac{x}{\sin X} = 2R$. For side $AC$ and angle $B$: $$ \frac{AC}{\sin B} = 2R $$ Substitute $AC=2c$ and $\sin B = \sin 120^\circ = \frac{\sqrt{3}}{2}$: $$ \frac{2c}{\sqrt{3}/2} = 2R $$ $$ \frac{4c}{\sqrt{3}} = 2R $$ $$ R = \frac{2c}{\sqrt{3}} $$ The ratio of the radius of the circumscribing circle to the side length $AB$ (which is $c$) is: $$ \frac{R}{c} = \frac{2c/\sqrt{3}}{c} = \frac{2}{\sqrt{3}} $$ Option 2 is correct. Step 7: Calculate the ratio of the area of $\triangle ABC$ to the area of the circumscribing circle and confirm Option 3. First, calculate the area of $\triangle ABC$. $$ \text{Area}(\triangle ABC) = \frac{1}{2} (AB)(AC)\sin A $$ We need $\sin A = \sin(60^\circ - C)$. Using the angle subtraction formula $\sin(X - Y) = \sin X \cos Y - \cos X \sin Y$: $$ \sin A = \sin 60^\circ \cos C - \cos 60^\circ \sin C $$ Substitute the values: $$ \sin A = \left(\frac{\sqrt{3}}{2}\right)\left(\frac{\sqrt{13}}{4}\right) - \left(\frac{1}{2}\right)\left(\frac{\sqrt{3}}{4}\right) $$ $$ \sin A = \frac{\sqrt{3}\sqrt{13}}{8} - \frac{\sqrt{3}}{8} = \frac{\sqrt{3}(\sqrt{13}-1)}{8} $$ Now, substitute into the area formula for $\triangle ABC$: $$ \text{Area}(\triangle ABC) = \frac{1}{2} (c)(2c)\left(\frac{\sqrt{3}(\sqrt{13}-1)}{8}\right) $$ $$ \text{Area}(\triangle ABC) = c^2 \frac{\sqrt{3}(\sqrt{13}-1)}{8} $$ Next, calculate the area of the circumscribing circle. The area of a circle is $\pi R^2$. We found $R = \frac{2c}{\sqrt{3}}$. $$ \text{Area(circle)} = \pi \left(\frac{2c}{\sqrt{3}}\right)^2 = \pi \left(\frac{4c^2}{3}\right) = \frac{4\pi c^2}{3} $$ Finally, find the ratio of the area of $\triangle ABC$ to the area of the circumscribing circle: $$ \frac{\text{Area}(\triangle ABC)}{\text{Area(circle)}} = \frac{c^2 \frac{\sqrt{3}(\sqrt{13}-1)}{8}}{\frac{4\pi c^2}{3}} $$ $$ \frac{\text{Area}(\triangle ABC)}{\text{Area(circle)}} = \frac{\sqrt{3}(\sqrt{13}-1)}{8} \cdot \frac{3}{4\pi} $$ $$ \frac{\text{Area}(\triangle ABC)}{\text{Area(circle)}} = \frac{3\sqrt{3}(\sqrt{13}-1)}{32\pi} $$ Option 3 is correct. Step 8: Calculate the ratio of the sides $AB$ to $AC$ and confirm Option 4. From Step 1, we established that $AB=c$ and $AC=2c$. Therefore, the ratio $\frac{AB}{AC}$ is: $$ \frac{AB}{AC} = \frac{c}{2c} = \frac{1}{2} $$ Option 4 is correct. Step 9: Conclusion. All options provided are correct based on our derivations. Option 1: The value of ratio $\frac{BC}{AB}$ is equal to $\frac{\sqrt{13}-1}{2}$. Option 2: The value of the ratio of radius of the circle circumscribing the triangle $ABC$ to the side length $AB$ is equal to $\frac{2}{\sqrt{3}}$. Option 3: The ratio of the area of $\triangle ABC$ to the area of the circle circumscribing $\triangle ABC$ is equal to $\frac{3\sqrt{3}}{32\pi}(\sqrt{13}-1)$. Option 4: The value of ratio of the sides $AB$ to $AC$ is equal to $1/2$. The final answer is $\boxed{\text{All options are correct}}$.