Cevian Theorem
Stewart's Theorem
A theorem relating the length of a cevian to the lengths of the sides of a triangle. This theorem is easily proven using the law of cosines.
See also
Key Formula
b2⋅m+c2⋅n−d2⋅a=a⋅m⋅n
Where:
- a = The length of the side of the triangle on which the cevian lands (the full side opposite vertex A), so that a = m + n
- b = The length of the side of the triangle adjacent to segment n (side AC)
- c = The length of the side of the triangle adjacent to segment m (side AB)
- d = The length of the cevian from vertex A to the opposite side
- m = The length of the segment of side a between vertex B and the foot of the cevian (adjacent to side c)
- n = The length of the segment of side a between the foot of the cevian and vertex C (adjacent to side b)
Worked Example
Problem: In triangle ABC, side BC = 10, side AC = 8, and side AB = 6. A cevian AD is drawn from vertex A to point D on side BC such that BD = 4 and DC = 6. Find the length of cevian AD.
Step 1: Identify all the values. Here a = BC = 10, b = AC = 8, c = AB = 6, m = BD = 4, n = DC = 6, and d = AD is unknown. Verify that m + n = a: 4 + 6 = 10. ✓
a=10,b=8,c=6,m=4,n=6
Step 2: Write Stewart's Theorem and substitute the known values.
b2⋅m+c2⋅n−d2⋅a=a⋅m⋅n
Step 3: Compute each term on the left side (excluding d² · a) and the right side.
82⋅4+62⋅6−d2⋅10=10⋅4⋅6
Step 4: Simplify the arithmetic: 64 · 4 = 256, 36 · 6 = 216, and the right side is 240.
256+216−10d2=240
Step 5: Solve for d². Combine: 472 − 10d² = 240, so 10d² = 232, giving d² = 23.2. Then take the square root.
d=23.2=5116=5229=52145≈4.817
Answer: The cevian AD has length 52145≈4.82.
Another Example
This example applies the Cevian Theorem to a median, where m = n = a/2. When the cevian bisects the opposite side, Stewart's Theorem reduces to the median length formula. This shows how Stewart's Theorem generalizes several well-known results.
Problem: In triangle ABC, AB = 5, AC = 7, and BC = 8. M is the midpoint of BC. Use the Cevian Theorem to find the length of the median AM.
Step 1: Since M is the midpoint of BC, we have m = BM = 4 and n = MC = 4, with a = BC = 8. Also c = AB = 5 and b = AC = 7.
a=8,b=7,c=5,m=4,n=4
Step 2: Substitute into Stewart's Theorem.
72⋅4+52⋅4−d2⋅8=8⋅4⋅4
Step 3: Compute: 49 · 4 = 196, 25 · 4 = 100, and the right side is 128.
196+100−8d2=128
Step 4: Solve: 296 − 8d² = 128, so 8d² = 168, giving d² = 21.
d=21≈4.583
Answer: The median AM has length 21≈4.58.
Frequently Asked Questions
What is the difference between the Cevian Theorem (Stewart's Theorem) and Ceva's Theorem?
Stewart's Theorem (the Cevian Theorem) calculates the length of a single cevian given the triangle's side lengths and the segments it creates. Ceva's Theorem, by contrast, deals with three cevians at once—it gives a condition for when three cevians are concurrent (all pass through a single point). Stewart's Theorem is a metric formula; Ceva's Theorem is a concurrency criterion.
When do you use the Cevian Theorem?
Use it whenever you know the three side lengths of a triangle and the position where a line from a vertex meets the opposite side, and you need the length of that line segment. It works for medians, angle bisectors, altitudes, or any arbitrary cevian. It is especially useful when you do not know any angles.
How do you prove Stewart's Theorem?
The standard proof applies the Law of Cosines twice. In triangle ABD, express cos(∠ADB) in terms of c, m, and d. In triangle ADC, express cos(∠ADC) using b, n, and d. Since ∠ADB and ∠ADC are supplementary, their cosines are negatives of each other. Setting up this relationship and eliminating the cosine term yields Stewart's formula.
Cevian Theorem (Stewart's Theorem) vs. Ceva's Theorem
| Cevian Theorem (Stewart's Theorem) | Ceva's Theorem | |
|---|---|---|
| What it finds | The length of a single cevian | Whether three cevians meet at one point |
| Formula type | Metric (gives a numerical length) | Ratio condition: (BD/DC)(CE/EA)(AF/FB) = 1 |
| Number of cevians involved | One | Three |
| Typical use case | Finding the length of a median, angle bisector, or arbitrary cevian | Proving that altitudes, medians, or other triples of cevians are concurrent |
Why It Matters
Stewart's Theorem appears frequently in competition mathematics and standardized geometry problems whenever you need to compute a cevian length without knowing any angles. It directly generalizes the median length formula and, combined with the angle bisector length formula, unifies several key triangle results. Mastering it gives you a powerful tool for any problem involving internal segments of a triangle.
Common Mistakes
Mistake: Mixing up which segment (m or n) is paired with which side (b or c).
Correction: Remember that m is the segment adjacent to vertex B (and thus adjacent to side c = AB), while n is adjacent to vertex C (and side b = AC). The pairing is: b² · m + c² · n, not b² · n + c² · m. A mnematch here flips the answer.
Mistake: Forgetting the sign convention — writing the formula as b²m + c²n + d²a = amn instead of subtracting d²a.
Correction: The correct form subtracts the cevian term: b²m + c²n − d²a = amn. Some textbooks rearrange this as b²m + c²n = a(d² + mn), which is equivalent. Double-check which form your source uses before substituting.
Related Terms
- Cevian — The line segment whose length this theorem computes
- Theorem — General classification of this mathematical result
- Triangle — The geometric figure in which cevians are drawn
- Law of Cosines — Used in the standard proof of Stewart's Theorem
- Ceva's Theorem — Related theorem about concurrency of three cevians
- Menelaus's Theorem — Related theorem about collinearity involving a transversal
- Side of a Polygon — The triangle sides that appear in the formula