Triangle Centers & Cevians

What you need to know

• Angle bisector theorem: the bisector from $A$ meets $BC$ at $D$ with $\frac{BD}{DC} = \frac{AB}{AC}$. The bisectors meet at the incenter $I$, and $\angle BIC = 90^\circ + \frac12\angle A$.
• Inradius $r = \dfrac{\text{Area}}{s}$ and circumradius $R = \dfrac{abc}{4\cdot\text{Area}}$, with $s$ the semiperimeter and the area usually from Heron's formula.
• The centroid divides each median $2:1$ from the vertex, and the medians cut the triangle into six equal-area triangles.
• Stewart's theorem for cevian $AD$ with $BD = m$, $DC = n$, $BC = a$: $b^2 m + c^2 n = a(d^2 + mn)$; for an angle bisector, $AD^2 = AB\cdot AC - BD\cdot DC$.
• Tangent lengths from the vertices to the incircle are $s-a$, $s-b$, $s-c$.

How AMC 10 tests it

• "Triangle with sides $13, 14, 15$ (or $5, 12, 13$): find the inradius or circumradius." Compute the area first.
• An angle bisector meets the opposite side; find a segment or the area of one piece (areas are in ratio $AB:AC$).
• Problems 18–24: a cevian length via Stewart or the law of cosines, or incircle tangent lengths locating a tangency point.

Standard approaches

1. Compute the area (Heron or a known altitude), then convert to $r$ or $R$.
2. For a bisector, split the opposite side with the bisector theorem, then use area ratios or Stewart.
3. For medians, use the centroid's $2:1$ ratio and equal-area sub-triangles.
4. Whenever an incircle appears, write the tangent lengths $s-a, s-b, s-c$ on the figure.

Worked example

In triangle $ABC$, $AB = 8$, $AC = 6$, and $BC = 7$. The bisector of $\angle A$ meets $\overline{BC}$ at $D$. What is $AD$?

(A) $5$ (B) $\dfrac{11}{2}$ (C) $6$ (D) $\dfrac{13}{2}$ (E) $7$

Solution. By the angle bisector theorem, $BD:DC = 8:6 = 4:3$, so $BD = 4$ and $DC = 3$. Stewart's theorem gives
$$6^2\cdot 4 + 8^2\cdot 3 = 7\left(AD^2 + 4\cdot 3\right) \implies 336 = 7AD^2 + 84,$$
so $AD^2 = 36$ and $AD = 6$. The answer is $\boxed{\textbf{(C)}\ 6}$.

Pitfalls

• Using the full perimeter instead of the semiperimeter in $r = A/s$ or Heron's formula.
• Assuming the bisector, median, and altitude from a vertex coincide; they do only at the apex of an isosceles triangle.
• Flipping the centroid ratio to $1:2$ from the vertex.

Traps that recur

• Forgetting that triangle ADE has one quarter of the area (midpoints halve both sides) or applying the bisector ratio 5:1 to the areas the wrong way round. (2018 AMC 10A #24)
• Assuming the two lines are medians or that the quadrilateral is a parallelogram, or guessing the answer from 3 + 7 + 7 and a total of 34 or 35. (2006 AMC 10B #23)
• Using masses from the two cevians on inconsistent scales (T must receive the same total from both), or reading off BD/CD = 4/11 and inverting the requested ratio. (2004 AMC 10B #20)
• Placing the circumcenter somewhere other than the midpoint of the hypotenuse, or computing r as area over perimeter (giving 1) instead of area over semiperimeter. (2004 AMC 10B #22)