Triangle Centers & Cevians

What you need to know

In triangle $ABC$, let $a = BC$, $b = CA$, $c = AB$ (each side is named for the corner across from it), and let $s = \frac{a+b+c}{2}$, the semiperimeter.

• Angle Bisector Theorem. If the bisector of $\angle A$ meets $BC$ at $D$, then $\frac{BD}{DC} = \frac{AB}{AC}$. Example: $AB = 6$, $AC = 9$, $BC = 10$ give $BD = 4$ and $DC = 6$.
• Incenter. The angle bisectors meet at the incenter $I$, the center of the inscribed circle; $\angle BIC = 90^\circ + \frac12\angle A$. Example: if $\angle A = 60^\circ$, then $\angle BIC = 120^\circ$.
• Inradius and circumradius. The inscribed circle has radius $r = \dfrac{\text{Area}}{s}$. The circle through the three corners has radius $R = \dfrac{abc}{4\cdot\text{Area}}$. The area usually comes from Heron's Formula. Example: the $13$-$14$-$15$ triangle has $s = 21$ and area $84$, so $r = 4$ and $R = \frac{13\cdot14\cdot15}{4\cdot84} = \frac{65}{8}$.
• Centroid. The medians (corner to midpoint of the opposite side) meet at the centroid, which divides each median $2:1$, longer part next to the corner. They cut the triangle into six triangles of equal area. Example: a median of length $9$ splits into $6$ and $3$.
• Stewart's Theorem. A cevian is a segment from a corner to the opposite side. For cevian $AD$ of length $d$, with $BD = m$ and $DC = n$: $b^2 m + c^2 n = a(d^2 + mn)$. For an angle bisector there is a shortcut: $AD^2 = AB\cdot AC - BD\cdot DC$.
• Tangent lengths. The two tangent segments from $A$ to the inscribed circle have length $s-a$; from $B$, $s-b$; from $C$, $s-c$. Example: $a = 14$, $b = 15$, $c = 13$ give $s = 21$ and tangent lengths $7$, $6$, $8$.

How AMC 10 tests it

• "A triangle has sides $13, 14, 15$ (or $5, 12, 13$); find the inradius or circumradius." Find the area first.
• An angle bisector meets the opposite side; find a piece of it, or the area of one part (areas are in the ratio $AB:AC$).
• Problems 18–24: a cevian length from Stewart's Theorem or the Law of Cosines, or tangent lengths of the inscribed circle that locate a point of tangency.

Standard approaches

1. Find the area (Heron's Formula or a known height), then turn it into $r$ or $R$.
2. For a bisector, split the opposite side with the Angle Bisector Theorem, then use area ratios or Stewart's Theorem.
3. For medians, use the centroid's $2:1$ ratio and the six equal-area triangles.
4. Whenever there is an inscribed circle, 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. Step 1. By the Angle Bisector Theorem, $BD:DC = 8:6 = 4:3$. The pieces add up to $7$, so $BD = 4$ and $DC = 3$.

Step 2. In Stewart's Theorem, $a = 7$, $b = AC = 6$, $c = AB = 8$, $m = 4$, $n = 3$:
$$6^2\cdot 4 + 8^2\cdot 3 = 7\left(AD^2 + 4\cdot 3\right) \implies 336 = 7AD^2 + 84.$$

Step 3. So $7AD^2 = 252$, $AD^2 = 36$, and $AD = 6$.

The answer is $\boxed{\textbf{(C)}\ 6}$.

Pitfalls

• Using the whole perimeter instead of the semiperimeter in $r = \frac{\text{Area}}{s}$ or in Heron's Formula. Example: the $3$-$4$-$5$ triangle has $s = 6$ (not $12$), so $r = \frac66 = 1$.
• Assuming the bisector, median, and altitude from a corner are the same segment. They are only from the corner between the two equal sides of an isosceles triangle.
• Flipping the centroid's ratio to $1:2$ from the corner. The longer piece touches the corner.

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)