📏 Special Segments in Triangles

Special segments in triangles are fundamental tools in AMC geometry. Each has unique properties that frequently appear in contest problems.

🎯 Key Concepts

Medians

Definition: Line segment from vertex to midpoint of opposite side Properties:

  • Three medians meet at centroid
  • Centroid divides each median in ratio 2:1
  • Medians divide triangle into six equal areas
  • Length formula: $m_a = \frac{1}{2}\sqrt{2b^2 + 2c^2 - a^2}$

Altitudes

Definition: Perpendicular line segment from vertex to opposite side (or extension) Properties:

  • Three altitudes meet at orthocenter
  • In right triangles, orthocenter is the right angle vertex
  • Area formula: $A = \frac{1}{2} \times \text{base} \times \text{height}$

Angle Bisectors

Definition: Line segment that bisects an angle Properties:

  • Three angle bisectors meet at incenter
  • Incenter is equidistant from all three sides
  • Angle Bisector Theorem: $\frac{BD}{DC} = \frac{AB}{AC}$
  • Length formula: $t_a = \frac{2bc}{b+c}\cos\frac{A}{2}$

Midsegments

Definition: Line segment connecting midpoints of two sides Properties:

  • Parallel to third side
  • Half the length of third side
  • Divides triangle into four congruent triangles

🔍 Micro-Examples

Median Example

In triangle with sides 3, 4, 5, the median to side 5 is $m = \frac{1}{2}\sqrt{2(3^2) + 2(4^2) - 5^2} = \frac{1}{2}\sqrt{18 + 32 - 25} = \frac{1}{2}\sqrt{25} = \frac{5}{2}$.

Altitude Example

In right triangle with legs 3 and 4, the altitude to hypotenuse is $h = \frac{3 \cdot 4}{5} = \frac{12}{5}$.

Angle Bisector Example

In triangle with sides 3, 4, 5, if angle bisector divides opposite side in ratio 3:4, then the two segments are $\frac{3 \cdot 5}{7} = \frac{15}{7}$ and $\frac{4 \cdot 5}{7} = \frac{20}{7}$.

Midsegment Example

In triangle with sides 3, 4, 5, the midsegment parallel to side 5 has length $\frac{5}{2} = 2.5$.

⚠️ Common Traps

Pitfall: Confusing median and altitude

  • Fix: Median goes to midpoint, altitude goes perpendicular

Pitfall: Wrong angle bisector theorem setup

  • Fix: Remember $\frac{BD}{DC} = \frac{AB}{AC}$

Pitfall: Forgetting midsegment properties

  • Fix: Midsegment is parallel and half the length

Pitfall: Wrong altitude in right triangles

  • Fix: In right triangles, altitude to hypotenuse is special

🎯 AMC-Style Worked Example

Problem: In triangle $ABC$, $AB = 6$, $AC = 8$, and $BC = 10$. Point $D$ is the midpoint of $BC$. Find the length of median $AD$.

Solution: Using the median length formula: $m_a = \frac{1}{2}\sqrt{2b^2 + 2c^2 - a^2}$

Where $a = BC = 10$, $b = AC = 8$, $c = AB = 6$: $m_a = \frac{1}{2}\sqrt{2(8^2) + 2(6^2) - 10^2}$ $m_a = \frac{1}{2}\sqrt{2(64) + 2(36) - 100}$ $m_a = \frac{1}{2}\sqrt{128 + 72 - 100}$ $m_a = \frac{1}{2}\sqrt{100}$ $m_a = \frac{1}{2} \cdot 10 = 5$

Answer: $AD = 5$

🔗 Related Topics

💡 Quick Reference

Segment Properties

  • Medians: Meet at centroid, divide in 2:1 ratio
  • Altitudes: Meet at orthocenter, perpendicular to sides
  • Angle Bisectors: Meet at incenter, equidistant from sides
  • Midsegments: Parallel to third side, half the length

Length Formulas

  • Median: $m_a = \frac{1}{2}\sqrt{2b^2 + 2c^2 - a^2}$
  • Altitude: $h_a = \frac{2A}{a}$ (where $A$ is area)
  • Angle Bisector: $t_a = \frac{2bc}{b+c}\cos\frac{A}{2}$

Special Cases

  • Right triangles: Altitude to hypotenuse is special
  • Isosceles triangles: Altitude, median, and angle bisector coincide
  • Equilateral triangles: All segments are equal

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