GeometryDifficulty 45-907 tagged problems

Stewart's Theorem

Stewart's Theorem is the habit of relating a cevian's length to the side lengths it cuts. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.

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The Key Result

$$b^2 n + c^2 m = a\,(d^2 + mn) \qquad (\text{man} + \text{dad} = \text{bmb} + \text{cnc}),\ \ m = DC,\ n = BD$$

Stewart's theorem gives the length of a cevian in terms of the triangle's three sides and the two pieces it cuts the opposite side into.

Why It Works

1
In $\triangle ABC$, let the cevian $AD = d$ hit side $BC = a$ at $D$, splitting it into $BD = n$ (next to $B$, where $AB = c$) and $DC = m$ (next to $C$, where $AC = b$), so $m + n = a$.
2
Apply the Law of Cosines in $\triangle ADC$ to angle $\angle ADC$: $b^2 = d^2 + m^2 - 2dm\cos\angle ADC$.
3
Apply it in $\triangle ADB$ to the supplementary angle $\angle ADB = 180^\circ - \angle ADC$, whose cosine is $-\cos\angle ADC$: $c^2 = d^2 + n^2 + 2dn\cos\angle ADC$.
4
Eliminate the cosine: multiply the first equation by $n$, the second by $m$, and add. The $\cos$ terms cancel ($-2dmn\cos + 2dmn\cos = 0$), leaving $b^2 n + c^2 m = (d^2)(m+n) + m^2 n + n^2 m$.
5
Since $m + n = a$ and $m^2 n + n^2 m = mn(m+n) = amn$, this is $b^2 n + c^2 m = a d^2 + a mn = a(d^2 + mn)$ — Stewart's theorem (with $m,n$ matched to $b,c$ as set up; the mnemonic 'man + dad = bmb + cnc' encodes it).

Worked Examples

Example 1

In $\triangle ABC$, $AB = 4$, $AC = 7$, and $BC = 9$. A cevian $AD$ meets $BC$ at $D$ with $BD = 6$ and $DC = 3$. Find the length $AD$.

1
Set up Stewart's with $a = BC = 9$, $b = AC = 7$, $c = AB = 4$, $m = DC = 3$ (next to $C$), $n = BD = 6$ (next to $B$), and $d = AD$.
2
Plug into $b^2 n + c^2 m = a(d^2 + mn)$ — note each side² multiplies the segment at the OPPOSITE end: $7^2\cdot 6 + 4^2\cdot 3 = 9\,(d^2 + 6\cdot 3)$.
3
Left side: $49\cdot 6 + 16\cdot 3 = 294 + 48 = 342$. So $342 = 9(d^2 + 18)$, giving $d^2 + 18 = 38$.
4
Thus $d^2 = 20$ and $d = 2\sqrt5$.

Answer:

$AD = 2\sqrt5$

Warm-up

In $\triangle ABC$, $AB = 5$, $AC = 7$, $BC = 8$. Find the length of the median $AD$ from $A$ to the midpoint $D$ of $BC$.

1
A median bisects $BC$, so $m = DC = 4$ and $n = BD = 4$, with $a = BC = 8$, $b = AC = 7$, $c = AB = 5$.
2
Stewart's: $b^2 n + c^2 m = a(d^2 + mn)$, i.e. $49\cdot 4 + 25\cdot 4 = 8(d^2 + 16)$.
3
Left side: $196 + 100 = 296$, so $296 = 8(d^2 + 16)$, giving $d^2 + 16 = 37$ and $d^2 = 21$.
4
Therefore $AD = \sqrt{21}$. (Check with the median formula $m_a^2 = \tfrac{2b^2 + 2c^2 - a^2}{4} = \tfrac{98 + 50 - 64}{4} = \tfrac{84}{4} = 21$.)

Answer:

$AD = \sqrt{21}$

Contest level

In $\triangle ABC$, $AB = 4$, $AC = 6$, $BC = 5$. The internal bisector of $\angle A$ meets $BC$ at $D$. Find the length of the bisector $AD$.

1
By the angle-bisector theorem $\dfrac{BD}{DC} = \dfrac{AB}{AC} = \dfrac{4}{6} = \dfrac{2}{3}$, so $BD = \tfrac25\cdot 5 = 2$ and $DC = \tfrac35\cdot 5 = 3$. Thus $n = BD = 2$, $m = DC = 3$.
2
Stewart's with $a = 5$, $b = AC = 6$, $c = AB = 4$: $b^2 n + c^2 m = a(d^2 + mn)$, i.e. $36\cdot 2 + 16\cdot 3 = 5(d^2 + 2\cdot 3)$.
3
Left side: $72 + 48 = 120$, so $120 = 5(d^2 + 6)$, giving $d^2 + 6 = 24$ and $d^2 = 18$.
4
Therefore $AD = \sqrt{18} = 3\sqrt2$. (Check with the bisector-length formula $t_a^2 = bc\left[1 - \left(\tfrac{a}{b+c}\right)^2\right] = 24\left[1 - \tfrac{25}{100}\right] = 24\cdot\tfrac34 = 18$.)

Answer:

$AD = 3\sqrt2$

Olympiad / Challenge

Derive Apollonius' median formula $m_a^2 = \dfrac{2b^2 + 2c^2 - a^2}{4}$ as the special case of Stewart's theorem when the cevian is a median.

1
For a median, $D$ is the midpoint of $BC$, so $m = n = \dfrac{a}{2}$ and $mn = \dfrac{a^2}{4}$.
2
Stewart's $b^2 n + c^2 m = a(d^2 + mn)$ becomes $b^2\cdot\dfrac{a}{2} + c^2\cdot\dfrac{a}{2} = a\left(m_a^2 + \dfrac{a^2}{4}\right)$.
3
Divide both sides by $a$: $\dfrac{b^2 + c^2}{2} = m_a^2 + \dfrac{a^2}{4}$.
4
Solve for $m_a^2$: $m_a^2 = \dfrac{b^2 + c^2}{2} - \dfrac{a^2}{4} = \dfrac{2b^2 + 2c^2 - a^2}{4}$.

Answer:

$m_a^2 = \dfrac{2b^2 + 2c^2 - a^2}{4}$ (Stewart's with $m = n = a/2$).

Going Deeper

Generalization: Stewart's theorem is two Law-of-Cosines applications on the supplementary angles $\angle ADB$ and $\angle ADC$, fused so the cosine cancels — so it is really 'the Law of Cosines for a cevian.' Specializing the foot gives the median formula (foot at the midpoint) and combines with the angle-bisector ratio to give the bisector-length formula.
Where it appears: any AMC/AIME problem that gives all three sides AND a cevian's foot position (a median, an angle bisector, or a stated ratio) and asks for the cevian's length is a one-line Stewart computation. It is the standard alternative to placing coordinates.
Pitfall: the side-to-segment pairing is the universal trap. Each side SQUARED multiplies the segment at the segment's FAR (opposite) end: $b^2$ (= $AC^2$, side at $C$) multiplies $BD$ (the segment at $B$). Pairing a side with the adjacent segment instead — the seductive misreading of the 'bmb + cnc' mnemonic — gives a wrong length, as the corrected worked example shows.

Spot the Signal

Use it when you know the side lengths and a cevian and need its length, or the reverse.
You can describe the hard part as relating a cevian's length to the side lengths it cuts, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.

Learn the Move

Start by labeling the cevian and its two sub-segments, then applying $man + dad = bmb + cnc$.
Name the relation that makes Stewart's Theorem legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.

Avoid These Traps

Mismatching which sub-segment is $m$ versus $n$, or which side multiplies which.
Applying Stewart's Theorem because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Label the pieces once, carefully; then the formula does the rest.

Try it on:

Find a cevian or segment length in a triangle from the side lengths.