Honors Geometry: Bisectors, Medians, and Altitudes of Triangles

Fundamental Geometric Constructions and Definitions

• Perpendicular Bisector: Defined as the line that is perpendicular to a segment at exactly its midpoint. It divides the segment into two congruent halves while forming a $90^{\circ}$ angle.

• Equidistant: A term used when a point is the same distance from two or more points or objects. This concept is central to the theorems regarding bisectors.

• Midpoint and Slope Calculation: For a segment with endpoints $(2, 8)$ and $(-4, 6)$, the following foundational calculations are used:

  • Midpoint Formula: $(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$. For the given values: $(\frac{2 + (-4)}{2}, \frac{8 + 6}{2}) = (-1, 7)$.

  • Slope Formula: $m = \frac{y_2 - y_1}{x_2 - x_1}$. For the given values: $m = \frac{6 - 8}{-4 - 2} = \frac{-2}{-6} = \frac{1}{3}$.

Perpendicular Bisectors: Theorems and Converses

• Perpendicular Bisector Theorem: If a point is located on the perpendicular bisector of a segment, then it is equidistant from the endpoints of that segment.

  • Example Representation: If $XY \perp AB$ at the midpoint of $AB$, and point $Y$ is on the bisector, then the distance $YA$ must equal the distance $YB$ ($YA = YB$).

• Converse of the Perpendicular Bisector Theorem: If a point is equidistant from the endpoints of a segment, then it is logically concluded that the point must lie on the perpendicular bisector of that segment.

  • Example Representation: If $XA = XB$, then point $X$ lies on the perpendicular bisector of segment $AB$.

Angle Bisectors: Theorems and Converses

• Angle Bisector Theorem: If a point is on the bisector of an angle, then it is equidistant from the sides of the angle. The distance to the side is always measured along a perpendicular segment from the point to the side.

  • Condition: Given $\angle APC \cong \angle BPC$, one can conclude the distances from $P$ to sides $A$ and $B$ are equal.

• Converse of the Angle Bisector Theorem: If a point in the interior of an angle is equidistant from the sides of the angle, then it is on the bisector of the angle.

  • Summary Logic: Given that the perpendicular segments from a point to the sides are congruent, we conclude the point lies on the angle bisector.

Coordinate Geometry: Equations of Perpendicular Bisectors

Writing the equation of a perpendicular bisector involves several specific steps. For endpoints $C(6, -5)$ and $D(10, 1)$, follow this procedure:

1. Find the Midpoint of $CD$:

   • $M = (\frac{6 + 10}{2}, \frac{-5 + 1}{2}) = (8, -2)$.

2. Find the Slope of $CD$:

   • $m = \frac{1 - (-5)}{10 - 6} = \frac{6}{4} = \frac{3}{2}$.

3. Find the Slope of the Perpendicular Bisector:

   • The perpendicular slope ($m_{\perp}$) is the negative reciprocal of the original slope.

   • $m_{\perp} = -\frac{2}{3}$.

4. Use Point-Slope Form to Write the Equation:

   • Point-Slope Form: $y - y_1 = m(x - x_1)$.

   • Equation: $y - (-2) = -\frac{2}{3}(x - 8)$ or $y + 2 = -\frac{2}{3}(x - 8)$.

Bisectors of Triangles: The Circumcenter and Incenter

• Concurrency: When three or more lines intersect at a single point.

• Point of Concurrency: The specific point where concurrent lines meet.

• Circumcenter:

  • Definition: The intersection (point of concurrency) of the perpendicular bisectors of a triangle.

  • Circumcenter Theorem: The circumcenter of a triangle is equidistant from the vertices of the triangle.

  • Circumscribed Circle: The circumcenter serves as the center of a circle that contains all the vertices of the polygon. This circle is said to be circumscribed about the triangle.

  • Locations:

    • Acute Triangle: Inside the triangle.

    • Obtuse Triangle: Outside the triangle.

    • Right Triangle: On the midpoint of the hypotenuse.

• Incenter:

  • Definition: The intersection of the angle bisectors of a triangle.

  • Incenter Theorem: The incenter of a triangle is equidistant from the sides of the triangle.

  • Inscribed Circle: The incenter is the center of a circle that is within the triangle and touches each side at exactly one point.

  • Location: The incenter is always located inside the triangle.

Medians and Altitudes: The Centroid and Orthocenter

• Median: A segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side.

• Centroid:

  • Definition: The point of concurrency where the medians of a triangle intersect.

  • Centroid Theorem: The centroid is located $\frac{2}{3}$ of the distance from each vertex to the midpoint of the opposite side.

  • Physical Property: The centroid is the center of gravity; a triangular region will balance perfectly on its centroid.

  • Location: Always inside the triangle.

• Altitude: A perpendicular segment from a vertex to the line containing the opposite side of a triangle. Unlike a median, it does not necessarily bisect the side.

• Orthocenter:

  • Definition: The point of concurrency where the altitudes of a triangle intersect.

  • Locations:

    • Acute Triangle: Inside the triangle.

    • Obtuse Triangle: Outside the triangle.

    • Right Triangle: At the vertex containing the right angle.

Step-by-Step Computational Procedures

Finding the Circumcenter of $\triangle HJK$ with $H(0, 0)$, $J(10, 0)$, and $K(0, 6)$:

1. Graph the Triangle: Plot points $(0, 0)$, $(10, 0)$, and $(0, 6)$.

2. Equations of Perpendicular Bisectors:

   • For side $HJ$ (horizontal on the x-axis): Midpoint is $(5, 0)$. The perpendicular bisector is the vertical line $x = 5$.

   • For side $HK$ (vertical on the y-axis): Midpoint is $(0, 3)$. The perpendicular bisector is the horizontal line $y = 3$.

3. Find the Intersection: The intersection of $x = 5$ and $y = 3$ is the point $(5, 3)$. This is the circumcenter.

Finding the Orthocenter of $\triangle ABC$ with $A(-3, 3)$, $B(3, 7)$, and $C(3, 0)$:

1. Graph the Triangle.

2. Find Altitude Equations:

   • Find the slope of one side, determine the perpendicular slope, and use the opposite vertex to write an equation.

   • For side $BC$: It is a vertical line ($x=3$) from $(3, 7)$ to $(3, 0)$. The altitude from $A$ to side $BC$ must be a horizontal line ($y = 3$).

3. Solve the System of Equations: Follow steps for a second altitude to find the meeting point.

Numerical Practice and Application Scenarios

• Numerical Example 1 (Centroid): In $\triangle LMN$, if $RL = 21$ (a median) and $S$ is the centroid, finding $LS$ involves the $\frac{2}{3}$ rule: $LS = \frac{2}{3} \times 21 = 14$. Thus, $SR = 7$.

• Numerical Example 2 (Centroid): If $SQ = 4$ and $S$ is the centroid on median $NQ$, where $Q$ is the midpoint and $N$ is the vertex: $SQ$ is $\frac{1}{3}$ of the total median length. Therefore, $NS = 2 \times 4 = 8$, and the total length $NQ = 12$.

• Defensive Basketball Strategy: A player stands on the bisector of $\angle ABC$. This is advantageous because the bisector is equidistant from the paths segment $\vec{BA}$ and $\vec{BC}$. This allows the defender to have the best possible positioning to intercept passes to either guard.

• Municipal Planning (The Lake Problem): Towns Ashton ($A$), Branford ($B$), and Clearview ($C$) want a boat for fireworks. To be equidistant from all three towns (the vertices), the boat must be placed at the circumcenter of the triangle formed by the towns.

• Playground Safety: A principal wants a swing set equidistant from three fences (the sides of a triangular playground). This project requires finding the incenter, as it is the point equidistant from the sides of the triangle.

Concept Summary Table

Questions & Discussion

1. Identifying Bisector Information: For a point $A$ to lie on the perpendicular bisector of $DB$, the diagram must show that $AD \cong AB$ or that the line containing $A$ is perpendicular to $DB$ at its midpoint. Without these markings, there is not enough information to conclude $A$ is on the bisector.

2. Variable Solving: Given $LH = LK$, $m\angle HJL = (3y + 19)^{\circ}$, and $m\angle LJK = (4y + 5)^{\circ}$, with $L$ on the angle bisector, we set the angles equal: $3y + 19 = 4y + 5$. Subtracting $3y$ from both sides gives $19 = y + 5$. Subtracting $5$ gives $y = 14$.

3. Isosceles Triangle Proof: Given $\triangle ABC$ is isosceles with base $AC$ and $BD$ bisects $\angle ABC$, one can prove $BD$ is an altitude because the angle bisector from the vertex of an isosceles triangle to the base is also the perpendicular bisector and the altitude of that base.