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Menelaus' Theorem

How to use
  1. ▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original figure. Learn more about the controls
  2. Drag DD along BCBC or FF along the extension of ABAB, or AA, BB or CC; the readout under the canvas shows the three ratios and their product. Learn more about why it holds
  3. Labels toggles the point names, Measures writes each marked angle's size beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene
Proof?A line cuts BC at D, CA at E and the extension of AB at F. Triangles on BD and DC with the apex F have the same height, so BD/DC = [FBD]/[FDC]; likewise CE/EA = [FCE]/[FEA] and, with the apex E, AF/FB = [EAF]/[EFB]. Multiply: [FEA] cancels, and the four areas left pair up along the transversal, [FBD]/[FBE] = FD/FE and [FCE]/[FCD] = FE/FD, so the product is 1. Drag D along BC or F along the extension of AB, or a vertex.Menelaus' theorem
Display
Drag D along BC, F along the extension of AB, or A, B or C
ABCDFE
BD/DC = 1.00, CE/EA = 0.25, AF/FB = 4.03, product = 1.00
Step 1 / 6

Step explanations

1A line across the triangle
Given
A line (the transversal) crosses the sides BC and CA of triangle ABC at D and E, and the extension of the third side AB at F. To prove: (BD/DC) · (CE/EA) · (AF/FB) = 1.
Drag D or F: the line swings, E follows. Learn more about the statement

Getting Started with the Visualizer
The Statement
Cutting the Side BC
Cutting the Side CA
Cutting the Line AB
Pairing the Areas
Why It Holds
Dragging the Figure
Consequences



Getting Started with the Visualizer

DemoA run through the proof player
Step 0 of 7
Step through the proof, then drag DD along the side BCBC or FF along the extension of ABAB to swing the line, or drag a vertex, and watch the three ratios change while their product stays at 1. The picture below is the player with its parts numbered; the other tabs list what can be dragged, what each control does and what the step log shows.
Opening scene: the controls
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1Proof — the theorem loaded in the player; the ? gives its idea in one line
2Display — Labels shows the point names, Measures writes each marked angle's size beside its arc
3Draggable points — D slides along BC, F along the extension of AB; A, B and C move freely; the hollow point E follows
4Readout — the line under the figure: the three ratios BD/DC, CE/EA, AF/FB and their product, live while you drag
5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original figure
6Speed and step counter — how fast Play advances, and which of the 6 steps is on screen
7Step explanations — every step reached so far, with its justification tag and a link to the section that explains it; click an entry to jump back to it
Proof complete: step 6 of 6
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1The transversal — the dashed line through D, E and F; FD and FE are the bases of the last two pairs
2The shaded triangles — FBD and FDC on BC, FCE and FEA on CA, EFB on AB: every ratio is a quotient of two of their areas
3F beyond B — the third point, on the extension of AB; AF is the long piece, FB the short one
4Justification tags — Given, Same height (five times), Multiplying
5Replay — at the last step Play becomes Replay and Next is off

The Statement

A line that is not parallel to any side of a triangle and misses its vertices crosses all three side lines. If it enters the triangle it crosses two sides between their ends and the third one on its extension; such a line is called a transversal. Menelaus' theorem says: if a line meets the sides BCBC, CACA of triangle ABCABC at DD, EE and the line ABAB at FF, then

BDDC⋅CEEA⋅AFFB=1\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = 1


The three fractions are the same ones as in Ceva's theorem, taken going round the triangle in one direction; there the three points come from three concurrent lines, here they lie on one line. With signed lengths the product is −1-1 for Menelaus and +1+1 for Ceva, which is how the two are told apart; the player uses plain lengths. The converse holds too: if the product is 1 and exactly one (or all three) of the points lies on an extension, the three points are collinear.

The proof here is by areas, like the one for Ceva. In the player DD slides along BCBC and FF along the extension of ABAB beyond BB; EE is where the line through them meets CACA. The readout under the canvas shows the three ratios and their product.
ABCDFE
Opening scene, frozen

The transversal through D, E and F; F beyond B on the extension of AB.


Cutting the Side BC

The triangles FBDFBD and FDCFDC have their bases BDBD and DCDC on the same line and the same apex FF, so they have the same height, and their areas (written with square brackets) are in the ratio of their bases:

[FBD][FDC]=BDDC\frac{[FBD]}{[FDC]} = \frac{BD}{DC}


This is the one fact the whole proof uses, the same one that drives the angle bisector theorem and Ceva: a ratio of lengths along a line is a ratio of areas of triangles standing on them with a common apex.
ABCDFE
Cutting the side BC, frozen

FBD and FDC shaded: [FBD]/[FDC] = BD/DC.


Cutting the Side CA

The same apex FF serves for the second side. The bases CECE and EAEA lie on the line CACA, so the triangles FCEFCE and FEAFEA have the same height from FF, and

[FCE][FEA]=CEEA\frac{[FCE]}{[FEA]} = \frac{CE}{EA}
ABCDFE
Cutting the side CA, frozen

FCE and FEA shaded: [FCE]/[FEA] = CE/EA.


Cutting the Line AB

The third pair of segments, AFAF and FBFB, lies on the line ABAB, and FF is one of their ends, so FF cannot be the apex this time. The point EE is: it lies on the transversal and not on ABAB. The triangles EAFEAF and EFBEFB have the same height from EE, so

[EAF][EFB]=AFFB\frac{[EAF]}{[EFB]} = \frac{AF}{FB}


Since FF lies beyond BB, the segment FBFB is the short piece and AFAF the long one; the ratio is larger than 1 in the opening figure.
ABCDFE
Cutting the line AB, frozen

EAF and EFB with the apex E: [EAF]/[EFB] = AF/FB.


Pairing the Areas

Multiply the three results:

BDDC⋅CEEA⋅AFFB=[FBD][FDC]⋅[FCE][FEA]⋅[EAF][EFB]\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = \frac{[FBD]}{[FDC]} \cdot \frac{[FCE]}{[FEA]} \cdot \frac{[EAF]}{[EFB]}


The area [FEA][FEA] is the same as [EAF][EAF], so it cancels. Four areas are left, and they pair up along the transversal itself. The triangles FBDFBD and FBEFBE stand on the bases FDFD and FEFE, which lie on the transversal, with the common apex BB; the triangles FCEFCE and FCDFCD stand on FEFE and FDFD with the common apex CC. Same height again:

[FBD][FBE]=FDFE,[FCE][FCD]=FEFD\frac{[FBD]}{[FBE]} = \frac{FD}{FE}, \qquad \frac{[FCE]}{[FCD]} = \frac{FE}{FD}


The player highlights the four triangles; FBEFBE is the one shaded at the previous step as EFBEFB.
ABCDFE
Pairing the areas, frozen

The four areas left, paired along the transversal: FD/FE and FE/FD.


Why It Holds

Putting the two pairs together,

BDDC⋅CEEA⋅AFFB=[FBD][FBE]⋅[FCE][FCD]=FDFE⋅FEFD=1\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = \frac{[FBD]}{[FBE]} \cdot \frac{[FCE]}{[FCD]} = \frac{FD}{FE} \cdot \frac{FE}{FD} = 1


The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, then same-height triangles five times (three along the sides, two along the transversal), and a multiplication. Nothing else is needed: no similar triangles, no parallels, no measurement.

The usual textbook proof instead draws the perpendiculars from AA, BB, CC to the transversal and uses three pairs of similar right triangles; the area proof avoids similarity, which is why it fits this section, where similarity has no page yet.
ABCDFE
The conclusion, frozen

The product is (FD/FE)(FE/FD) = 1.


Dragging the Figure

Drag DD along BCBC or FF along the extension of ABAB: the transversal swings, EE slides along CACA, the three ratios in the readout change, and their product stays at 1.001.00. Drag AA, BB or CC to reshape the triangle; DD and FF keep their positions along their lines and EE is recomputed. Pull FF far out and watch AF/FBAF/FB sink towards 1 while CE/EACE/EA shrinks to make up for it.

The player keeps FF beyond BB and DD on the side BCBC itself, which forces EE onto the side CACA itself: a line through one side and one extension always crosses the second side. The theorem also holds when all three points lie on extensions (the line misses the triangle), with the same proof; the player keeps to the case where the line enters the triangle. A drag that would carry FF or a vertex out of the frame, or bring two of the points too close, is refused.
ABCDFE
After dragging F, frozen

F pulled further out: AF/FB nearer 1, CE/EA smaller, the product still 1.


Consequences

• A test for collinearity. The converse is the theorem's main use: to show that three points on the side lines of a triangle lie on one line, compute the three ratios and check that the product is 1 with an odd number of the points on extensions. Many classical collinearities (the Simson line, the Pascal line, the line through the three external centres of similitude of three circles) are proved this way.
• The dual of Ceva. Ceva's theorem uses the same three ratios for three concurrent cevians; Menelaus uses them for three collinear points. The two proofs by areas share their first step and differ afterwards: Ceva subtracts areas around PP, Menelaus pairs them along the transversal.
• Transversal of a cevian. Applying Menelaus to the triangle cut off by a cevian gives the ratio in which a second cevian divides the first; for two medians this gives the 2:12 : 1 ratio at the centroid in one line.
• Signed form. With directed lengths along each side the product is −1-1, which records that an odd number of the points lie on extensions; the player, working with plain lengths, shows the value 1.
• Trigonometric form. Replacing each ratio by a ratio of sines of the angles at the opposite vertex gives a trigonometric Menelaus, parallel to the trigonometric Ceva.