1. **Problem statement:** Given an isosceles triangle $\triangle ABC$ with $|AB| = |AC|$, and $M$ as the midpoint of $BC$, prove that the median $AM$ bisects the angle $\angle A$ and is perpendicular to $BC$. 2. **Key definitions and formulas:** - Median: A line segment joining a vertex to the midpoint of the opposite side. - Angle bisector: A line that divides an angle into two equal parts. - Perpendicular lines: Two lines that intersect at a right angle ($90^\circ$). 3. **Step 1: Setup coordinates for clarity** Place $B$ and $C$ on the x-axis for simplicity. Let $B = (0,0)$ and $C = (2a,0)$ for some $a > 0$. Since $M$ is the midpoint of $BC$, $M = (a,0)$. 4. **Step 2: Coordinates of $A$** Since $|AB| = |AC|$, $A$ lies on the perpendicular bisector of $BC$. The perpendicular bisector of $BC$ is the vertical line $x = a$. Let $A = (a,h)$ where $h > 0$. 5. **Step 3: Show $AM$ bisects $\angle A$** Vectors $AB$ and $AC$ are: $$\vec{AB} = (0 - a, 0 - h) = (-a, -h)$$ $$\vec{AC} = (2a - a, 0 - h) = (a, -h)$$ 6. **Step 4: Vector $AM$** $$\vec{AM} = (a - a, 0 - h) = (0, -h)$$ 7. **Step 5: Check if $AM$ bisects $\angle A$** Calculate the angle between $\vec{AM}$ and $\vec{AB}$: $$\cos \theta_1 = \frac{\vec{AM} \cdot \vec{AB}}{|\vec{AM}||\vec{AB}|} = \frac{0 \times (-a) + (-h)(-h)}{h \sqrt{a^2 + h^2}} = \frac{h^2}{h \sqrt{a^2 + h^2}} = \frac{h}{\sqrt{a^2 + h^2}}$$ Calculate the angle between $\vec{AM}$ and $\vec{AC}$: $$\cos \theta_2 = \frac{\vec{AM} \cdot \vec{AC}}{|\vec{AM}||\vec{AC}|} = \frac{0 \times a + (-h)(-h)}{h \sqrt{a^2 + h^2}} = \frac{h}{\sqrt{a^2 + h^2}}$$ Since $\cos \theta_1 = \cos \theta_2$, angles $\theta_1$ and $\theta_2$ are equal, so $AM$ bisects $\angle A$. 8. **Step 6: Show $AM$ is perpendicular to $BC$** $BC$ lies along the x-axis, so its direction vector is $(2a,0)$. $AM$ has direction vector $(0,-h)$. The dot product: $$\vec{AM} \cdot \vec{BC} = 0 \times 2a + (-h) \times 0 = 0$$ Since the dot product is zero, $AM$ is perpendicular to $BC$. **Final answer:** The median $AM$ bisects the angle $\angle A$ and is perpendicular to $BC$ in the isosceles triangle $\triangle ABC$ where $|AB| = |AC|$.