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Cevian

Cevian

A line segment, ray, or line that extends from a vertex of a triangle to the opposite side (which may be extended). Medians, altitudes, and angle bisectors are all examples of cevians.

 

Triangle with a cevian line drawn from a vertex to the opposite side, labeled "cevian.

 

 

See also

Ceva’s Theorem, Menelaus’s Theorem, Stewart's Theorem

Key Formula

d2=a2⋅m+b2⋅nm+n−m⋅nd^2 = \frac{a^2 \cdot m + b^2 \cdot n}{m + n} - m \cdot n
Where:
  • dd = Length of the cevian from vertex to the opposite side
  • aa = Length of the side adjacent to the segment of length n
  • bb = Length of the side adjacent to the segment of length m
  • mm = Length of one segment of the divided opposite side (adjacent to side b)
  • nn = Length of the other segment of the divided opposite side (adjacent to side a)

Worked Example

Problem: In triangle ABC, a cevian AD is drawn from vertex A to point D on side BC. Given that AB = 5, AC = 7, BD = 3, and DC = 6, find the length of cevian AD using Stewart's Theorem (the Cevian Theorem).
Step 1: Identify the variables for Stewart's Theorem. Here, side BC has total length m + n = 3 + 6 = 9. We set m = BD = 3 (adjacent to side AB = b = 5) and n = DC = 6 (adjacent to side AC = a = 7).
m=3,n=6,a=7,b=5m = 3, \quad n = 6, \quad a = 7, \quad b = 5
Step 2: Write Stewart's Theorem formula for the cevian length d = AD.
d2=a2⋅m+b2⋅nm+n−m⋅nd^2 = \frac{a^2 \cdot m + b^2 \cdot n}{m + n} - m \cdot n
Step 3: Substitute the known values into the formula.
d2=72⋅3+52⋅63+6−3⋅6=49⋅3+25⋅69−18d^2 = \frac{7^2 \cdot 3 + 5^2 \cdot 6}{3 + 6} - 3 \cdot 6 = \frac{49 \cdot 3 + 25 \cdot 6}{9} - 18
Step 4: Compute the numerator and simplify.
d2=147+1509−18=2979−18=33−18=15d^2 = \frac{147 + 150}{9} - 18 = \frac{297}{9} - 18 = 33 - 18 = 15
Step 5: Take the square root to find d.
d=15≈3.87d = \sqrt{15} \approx 3.87
Answer: The length of cevian AD is √15 ≈ 3.87.

Another Example

This example uses Ceva's Theorem instead of Stewart's Theorem, showing how cevians interact when they are concurrent. It focuses on ratios rather than lengths.

Problem: In triangle ABC, cevians AD, BE, and CF are concurrent (they all meet at a single interior point). Given that BD = 2, DC = 4, CE = 3, and EA = 6, find the ratio AF : FB using Ceva's Theorem.
Step 1: State Ceva's Theorem: three cevians AD, BE, and CF are concurrent if and only if the product of the ratios equals 1.
BDDC⋅CEEA⋅AFFB=1\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1
Step 2: Substitute the known segment lengths.
24⋅36⋅AFFB=1\frac{2}{4} \cdot \frac{3}{6} \cdot \frac{AF}{FB} = 1
Step 3: Simplify the known ratios.
12⋅12⋅AFFB=1  ⟹  14⋅AFFB=1\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{AF}{FB} = 1 \implies \frac{1}{4} \cdot \frac{AF}{FB} = 1
Step 4: Solve for AF/FB.
AFFB=4\frac{AF}{FB} = 4
Answer: AF : FB = 4 : 1.

Frequently Asked Questions

What is the difference between a cevian and a median?
A median is a specific type of cevian that connects a vertex to the midpoint of the opposite side. A cevian is more general — it can connect a vertex to any point on the opposite side (or its extension). Every median is a cevian, but not every cevian is a median.
Is an altitude always a cevian?
An altitude from a vertex is a cevian when the foot of the perpendicular lies on the opposite side of the triangle. In an obtuse triangle, the altitude from an acute-angled vertex falls on the extension of the opposite side, so it is still considered a cevian in the extended sense. The key requirement is that the segment goes from a vertex toward the line containing the opposite side.
What is Ceva's Theorem and how does it relate to cevians?
Ceva's Theorem states that three cevians AD, BE, and CF of a triangle ABC are concurrent (meet at one point) if and only if (BD/DC) · (CE/EA) · (AF/FB) = 1. This theorem is the main tool for proving that three cevians pass through a common point, such as the centroid, incenter, or orthocenter.

Cevian vs. Median

CevianMedian
DefinitionSegment from a vertex to any point on the opposite sideSegment from a vertex to the midpoint of the opposite side
Divides opposite sideInto two segments of any ratioInto two equal segments
Concurrency theoremCeva's Theorem (general condition)All three medians always meet at the centroid
Length formulaStewart's Theorem (general)Apollonius's Theorem (special case of Stewart's)
ExamplesMedians, altitudes, angle bisectors, and any other such segmentOnly the three medians of a triangle

Why It Matters

Cevians appear throughout geometry competitions and standardized tests whenever you work with triangles. Understanding cevians ties together several major theorems — Stewart's Theorem for computing lengths, Ceva's Theorem for proving concurrency, and the angle bisector theorem as a special case. Mastering this concept gives you a unified framework for solving problems about medians, altitudes, angle bisectors, and more.

Common Mistakes

Mistake: Confusing which side length pairs with which segment in Stewart's Theorem.
Correction: In the formula d² = (a²m + b²n)/(m+n) − mn, side a is opposite vertex A but adjacent to segment n, and side b is adjacent to segment m. Always draw and label a diagram before substituting.
Mistake: Assuming a cevian must lie inside the triangle.
Correction: A cevian can extend to a point on the extension of the opposite side. For example, in an obtuse triangle, the altitude from certain vertices lands outside the triangle, but it is still a cevian in the broader definition.

Related Terms