GeometryDifficulty 45-9014 tagged problems

Ceva and Menelaus

Ceva and Menelaus is the habit of using product-of-ratios criteria for concurrent cevians and collinear points. 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.

Practice this technique Match my level

The Key Result

$$\text{Ceva (concurrent): } \frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB} = 1; \quad \text{Menelaus (collinear): } \frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB} = -1$$

A product of three side-ratios equals one when the cevians meet at a point and minus one (signed) when the three points are collinear.

Why It Works

1
Take $\triangle ABC$ with $D$ on $BC$, $E$ on $CA$, $F$ on $AB$.
2
Ceva: the three cevians $AD, BE, CF$ are concurrent iff $\dfrac{BD}{DC}\cdot\dfrac{CE}{EA}\cdot\dfrac{AF}{FB} = 1$ (provable by replacing each ratio with an area ratio of triangles sharing a vertex).
3
Menelaus: the three points are collinear iff the same product of SIGNED ratios equals $-1$ (use directed lengths; an unsigned reading gives $1$ but you must track the configuration).
4
The two theorems are duals: concurrency vs. collinearity, distinguished by the sign in the directed product.

Worked Examples

Example 1

In $\triangle ABC$, points $D \in BC$, $E \in CA$, $F \in AB$ satisfy $\dfrac{BD}{DC} = 2$ and $\dfrac{CE}{EA} = 3$. If cevians $AD$, $BE$, $CF$ are concurrent, find $\dfrac{AF}{FB}$.

1
By Ceva's theorem, $\dfrac{BD}{DC}\cdot\dfrac{CE}{EA}\cdot\dfrac{AF}{FB} = 1$.
2
Substitute the known ratios: $2 \cdot 3 \cdot \dfrac{AF}{FB} = 1$.
3
Solve: $\dfrac{AF}{FB} = \dfrac{1}{6}$.

Answer:

$\dfrac{AF}{FB} = \dfrac{1}{6}$

Warm-up

In $\triangle ABC$, a line crosses side $AB$ at $F$, side $BC$ at $D$, and the extension of side $CA$ at $E$. If $\dfrac{AF}{FB} = \dfrac{2}{3}$ and $\dfrac{BD}{DC} = \dfrac{4}{1}$, find $\dfrac{CE}{EA}$ by Menelaus.

1
Menelaus (unsigned magnitude): $\dfrac{AF}{FB}\cdot\dfrac{BD}{DC}\cdot\dfrac{CE}{EA} = 1$ for a transversal of $\triangle ABC$.
2
Substitute: $\dfrac{2}{3}\cdot\dfrac{4}{1}\cdot\dfrac{CE}{EA} = 1$, i.e. $\dfrac{8}{3}\cdot\dfrac{CE}{EA} = 1$.
3
Solve: $\dfrac{CE}{EA} = \dfrac{3}{8}$.

Answer:

$\dfrac{CE}{EA} = \dfrac{3}{8}$

Contest level

In $\triangle ABC$, $D$ on $BC$ with $BD:DC = 2:1$ and $E$ on $CA$ with $CE:EA = 1:1$ ($E$ is the midpoint). Line $DE$ meets line $AB$ at $F$. Find $\dfrac{AF}{FB}$.

1
Apply Menelaus to $\triangle ABC$ with transversal $D\text{-}E\text{-}F$: $\dfrac{BD}{DC}\cdot\dfrac{CE}{EA}\cdot\dfrac{AF}{FB} = 1$ (unsigned).
2
Substitute the knowns: $\dfrac{2}{1}\cdot\dfrac{1}{1}\cdot\dfrac{AF}{FB} = 1$, so $2\cdot\dfrac{AF}{FB} = 1$.
3
Thus $\dfrac{AF}{FB} = \dfrac{1}{2}$ in magnitude; because the transversal must hit one side externally, $F$ lies on the extension of $AB$ beyond $A$.

Answer:

$\dfrac{AF}{FB} = \dfrac{1}{2}$ (with $F$ outside segment $AB$, on the extension past $A$).

Olympiad / Challenge

Prove Ceva's theorem from area ratios: if cevians $AD, BE, CF$ of $\triangle ABC$ concur at $P$, then $\dfrac{BD}{DC}\cdot\dfrac{CE}{EA}\cdot\dfrac{AF}{FB} = 1$.

1
On side $BC$, the foot $D$ of cevian $AD$ gives $\dfrac{BD}{DC} = \dfrac{[\triangle ABD]}{[\triangle ACD]}$ and also $\dfrac{BD}{DC} = \dfrac{[\triangle PBD]}{[\triangle PCD]}$ (same-height ratios).
2
Subtracting the second area pair from the first (the triangles share the cevian) gives $\dfrac{BD}{DC} = \dfrac{[\triangle ABD]-[\triangle PBD]}{[\triangle ACD]-[\triangle PCD]} = \dfrac{[\triangle APB]}{[\triangle APC]}$.
3
By the same argument on the other two sides, $\dfrac{CE}{EA} = \dfrac{[\triangle BPC]}{[\triangle BPA]}$ and $\dfrac{AF}{FB} = \dfrac{[\triangle CPA]}{[\triangle CPB]}$.
4
Multiply the three: $\dfrac{[\triangle APB]}{[\triangle APC]}\cdot\dfrac{[\triangle BPC]}{[\triangle BPA]}\cdot\dfrac{[\triangle CPA]}{[\triangle CPB]} = 1$, since every area appears once in a numerator and once in a denominator.

Answer:

The product telescopes to $1$, proving Ceva's concurrency condition.

Going Deeper

Generalization: Ceva and Menelaus are the affine 'concurrency vs. collinearity' dual pair, and both extend to SIGNED (directed) ratios — the only honest way to handle feet that land outside their segments. The signed product is $+1$ for Ceva (concurrent cevians) and $-1$ for Menelaus (collinear points); the trig form of Ceva ($\prod \sin$-ratios $= 1$) covers configurations given by angles.
Where it appears: these are the go-to tools the moment a problem says 'three cevians concur' or 'three points are collinear', and they shortcut many AIME ratio computations. Menelaus is also the standard engine behind projective lemmas (the complete quadrilateral, harmonic divisions).
Pitfall: a Menelaus transversal must cross an EVEN number of sides externally (one or three of the three feet lie outside their segments). If you treat every ratio as positive you can get a spurious answer; track signs, or at least verify the configuration forces exactly one external foot.

Spot the Signal

Look for problems where the key step is using product-of-ratios criteria for concurrent cevians and collinear points.
You can describe the hard part as using product-of-ratios criteria for concurrent cevians and collinear points, 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 identify the diagram relation that calls for ceva and menelaus, then rewrite the givens around it.
Name the relation that makes Ceva and Menelaus 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

Do not use Ceva and Menelaus just because the surface looks familiar; verify the required condition first.
Applying Ceva and Menelaus because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let ceva and menelaus reveal the geometric configuration; then compute only what remains.

Try it on:

Practice a contest problem where the key step is using product-of-ratios criteria for concurrent cevians and collinear points.

Back to techniques

GeometryDifficulty 45-9014 tagged problems

Ceva and Menelaus

Practice this technique Match my level

The Key Result

$$\text{Ceva (concurrent): } \frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB} = 1; \quad \text{Menelaus (collinear): } \frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB} = -1$$

A product of three side-ratios equals one when the cevians meet at a point and minus one (signed) when the three points are collinear.

Why It Works

1
Take $\triangle ABC$ with $D$ on $BC$, $E$ on $CA$, $F$ on $AB$.
2
Ceva: the three cevians $AD, BE, CF$ are concurrent iff $\dfrac{BD}{DC}\cdot\dfrac{CE}{EA}\cdot\dfrac{AF}{FB} = 1$ (provable by replacing each ratio with an area ratio of triangles sharing a vertex).
3
Menelaus: the three points are collinear iff the same product of SIGNED ratios equals $-1$ (use directed lengths; an unsigned reading gives $1$ but you must track the configuration).
4
The two theorems are duals: concurrency vs. collinearity, distinguished by the sign in the directed product.

Worked Examples

Example 1

In $\triangle ABC$, points $D \in BC$, $E \in CA$, $F \in AB$ satisfy $\dfrac{BD}{DC} = 2$ and $\dfrac{CE}{EA} = 3$. If cevians $AD$, $BE$, $CF$ are concurrent, find $\dfrac{AF}{FB}$.

1
By Ceva's theorem, $\dfrac{BD}{DC}\cdot\dfrac{CE}{EA}\cdot\dfrac{AF}{FB} = 1$.
2
Substitute the known ratios: $2 \cdot 3 \cdot \dfrac{AF}{FB} = 1$.
3
Solve: $\dfrac{AF}{FB} = \dfrac{1}{6}$.

Answer:

$\dfrac{AF}{FB} = \dfrac{1}{6}$

Warm-up

1
Menelaus (unsigned magnitude): $\dfrac{AF}{FB}\cdot\dfrac{BD}{DC}\cdot\dfrac{CE}{EA} = 1$ for a transversal of $\triangle ABC$.
2
Substitute: $\dfrac{2}{3}\cdot\dfrac{4}{1}\cdot\dfrac{CE}{EA} = 1$, i.e. $\dfrac{8}{3}\cdot\dfrac{CE}{EA} = 1$.
3
Solve: $\dfrac{CE}{EA} = \dfrac{3}{8}$.

Answer:

$\dfrac{CE}{EA} = \dfrac{3}{8}$

Contest level

In $\triangle ABC$, $D$ on $BC$ with $BD:DC = 2:1$ and $E$ on $CA$ with $CE:EA = 1:1$ ($E$ is the midpoint). Line $DE$ meets line $AB$ at $F$. Find $\dfrac{AF}{FB}$.

1
Apply Menelaus to $\triangle ABC$ with transversal $D\text{-}E\text{-}F$: $\dfrac{BD}{DC}\cdot\dfrac{CE}{EA}\cdot\dfrac{AF}{FB} = 1$ (unsigned).
2
Substitute the knowns: $\dfrac{2}{1}\cdot\dfrac{1}{1}\cdot\dfrac{AF}{FB} = 1$, so $2\cdot\dfrac{AF}{FB} = 1$.
3
Thus $\dfrac{AF}{FB} = \dfrac{1}{2}$ in magnitude; because the transversal must hit one side externally, $F$ lies on the extension of $AB$ beyond $A$.

Answer:

$\dfrac{AF}{FB} = \dfrac{1}{2}$ (with $F$ outside segment $AB$, on the extension past $A$).

Olympiad / Challenge

Prove Ceva's theorem from area ratios: if cevians $AD, BE, CF$ of $\triangle ABC$ concur at $P$, then $\dfrac{BD}{DC}\cdot\dfrac{CE}{EA}\cdot\dfrac{AF}{FB} = 1$.

1
On side $BC$, the foot $D$ of cevian $AD$ gives $\dfrac{BD}{DC} = \dfrac{[\triangle ABD]}{[\triangle ACD]}$ and also $\dfrac{BD}{DC} = \dfrac{[\triangle PBD]}{[\triangle PCD]}$ (same-height ratios).
2
Subtracting the second area pair from the first (the triangles share the cevian) gives $\dfrac{BD}{DC} = \dfrac{[\triangle ABD]-[\triangle PBD]}{[\triangle ACD]-[\triangle PCD]} = \dfrac{[\triangle APB]}{[\triangle APC]}$.
3
By the same argument on the other two sides, $\dfrac{CE}{EA} = \dfrac{[\triangle BPC]}{[\triangle BPA]}$ and $\dfrac{AF}{FB} = \dfrac{[\triangle CPA]}{[\triangle CPB]}$.
4
Multiply the three: $\dfrac{[\triangle APB]}{[\triangle APC]}\cdot\dfrac{[\triangle BPC]}{[\triangle BPA]}\cdot\dfrac{[\triangle CPA]}{[\triangle CPB]} = 1$, since every area appears once in a numerator and once in a denominator.

Answer:

The product telescopes to $1$, proving Ceva's concurrency condition.

Going Deeper

Generalization: Ceva and Menelaus are the affine 'concurrency vs. collinearity' dual pair, and both extend to SIGNED (directed) ratios — the only honest way to handle feet that land outside their segments. The signed product is $+1$ for Ceva (concurrent cevians) and $-1$ for Menelaus (collinear points); the trig form of Ceva ($\prod \sin$-ratios $= 1$) covers configurations given by angles.
Where it appears: these are the go-to tools the moment a problem says 'three cevians concur' or 'three points are collinear', and they shortcut many AIME ratio computations. Menelaus is also the standard engine behind projective lemmas (the complete quadrilateral, harmonic divisions).
Pitfall: a Menelaus transversal must cross an EVEN number of sides externally (one or three of the three feet lie outside their segments). If you treat every ratio as positive you can get a spurious answer; track signs, or at least verify the configuration forces exactly one external foot.

Spot the Signal

Look for problems where the key step is using product-of-ratios criteria for concurrent cevians and collinear points.
You can describe the hard part as using product-of-ratios criteria for concurrent cevians and collinear points, 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 identify the diagram relation that calls for ceva and menelaus, then rewrite the givens around it.
Name the relation that makes Ceva and Menelaus 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

Do not use Ceva and Menelaus just because the surface looks familiar; verify the required condition first.
Applying Ceva and Menelaus because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.

MathGrit Coach Note

Let ceva and menelaus reveal the geometric configuration; then compute only what remains.

Try it on:

Practice a contest problem where the key step is using product-of-ratios criteria for concurrent cevians and collinear points.