Visual Tools
Calculators
Tables
Mathematical Keyboard
Converters
Other Tools

Ceva's 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 PP anywhere inside the triangle, 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?Triangles ABD and ADC have the same height from A, so BD/DC equals the ratio of their areas; the same with apex P gives BD/DC = [PBD]/[PDC]. Subtracting the small triangles from the large ones leaves BD/DC = [ABP]/[ACP]. The other two cevians give CE/EA = [BCP]/[BAP] and AF/FB = [CAP]/[CBP], and the three fractions multiply to 1. Drag P inside, or a vertex.Ceva's theorem
Display
Drag P inside the triangle, or A, B or C
ABCPDEF
BD/DC = 1.28, CE/EA = 0.92, AF/FB = 0.85, product = 1.00
Step 1 / 6

Step explanations

1Three cevians through one point
Given
In triangle ABC the segments AD, BE and CF from the vertices to the opposite sides (cevians) all pass through the point P. To prove: (BD/DC) · (CE/EA) · (AF/FB) = 1.
Drag P: the three feet slide along the sides. Learn more about the statement

Getting Started with the Visualizer
The Statement
Same Height, Same Ratio
The Same with Apex P
Subtracting the Areas
The Other Two Cevians
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 the point PP anywhere inside the triangle, or 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
1234567
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 — P moves anywhere inside the triangle, A, B and C move freely; the hollow feet D, E, F follow
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
12345
1The triangles around P — ABP, BCP and CAP: each cevian ratio is a quotient of two of their areas
2The small triangles — PBD and PDC, the first pair with apex P; subtracted from ABD and ADC
3The three feet — D, E, F, where the cevians through P land on the sides
4Justification tags — Given, Same height (twice), Subtracting, Same argument, Multiplying
5Replay — at the last step Play becomes Replay and Next is off

The Statement

A cevian is a segment from a vertex of a triangle to a point on the opposite side. Ceva's theorem says when three of them pass through one point: if ADAD, BEBE and CFCF are cevians of triangle ABCABC meeting at a point PP inside it, then

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


Each fraction is the ratio in which a cevian cuts the side it lands on, taken going round the triangle in one direction. The converse holds too: if the product is 1, the three cevians are concurrent. That converse is how the theorem is mostly used; the medians, for instance, cut each side in the ratio 1:11 : 1, so the product is 11 and they meet at a point.

The proof here is by areas, and uses nothing but the formula half base times height. In the player PP is free inside the triangle; DD, EE, FF are where the lines through PP meet the sides. The readout under the canvas shows the three ratios and their product.
ABCPDEF
Opening scene, frozen

The three cevians AD, BE, CF through P.


Same Height, Same Ratio

The cevian ADAD splits the triangle into ABDABD and ADCADC. These two triangles have their bases BDBD and DCDC on the same line and the same apex AA, so they have the same height, and their areas (written with square brackets) are in the ratio of their bases:

[ABD][ADC]=BDDC\frac{[ABD]}{[ADC]} = \frac{BD}{DC}


This is the same fact that drives the angle bisector theorem page: a ratio of lengths on a line becomes a ratio of areas of triangles standing on them.
ABCPDEF
Same height, same ratio, frozen

ABD and ADC stand on BD and DC with the apex A: [ABD]/[ADC] = BD/DC.


The Same with Apex P

The point PP lies on ADAD, so the same two bases BDBD and DCDC carry a second pair of triangles with apex PP: PBDPBD and PDCPDC, shaded in the player. They too have equal heights, so

[PBD][PDC]=BDDC\frac{[PBD]}{[PDC]} = \frac{BD}{DC}


Two different pairs of areas, the same ratio.
ABCPDEF
The same with apex P, frozen

PBD and PDC shaded: [PBD]/[PDC] = BD/DC too.


Subtracting the Areas

If two fractions are equal, a/b=c/da/b = c/d, then so is the fraction of their differences, (a−c)/(b−d)=a/b(a - c)/(b - d) = a/b. Apply this to the two pairs of areas. Because PP is on ADAD, the small triangle PBDPBD sits inside ABDABD and leaves ABPABP; likewise PDCPDC inside ADCADC leaves ACPACP:

BDDC=[ABD]−[PBD][ADC]−[PDC]=[ABP][ACP]\frac{BD}{DC} = \frac{[ABD] - [PBD]}{[ADC] - [PDC]} = \frac{[ABP]}{[ACP]}


So the cevian from AA cuts the opposite side in the ratio of the two areas on either side of it at PP. The segments BDBD and DCDC have disappeared; only the triangles around PP remain.
ABCPDEF
Subtracting the areas, frozen

ABP and ACP shaded: BD/DC = [ABP]/[ACP].


The Other Two Cevians

Nothing in the previous three steps was special to the vertex AA. Running the same argument from BB along the cevian BEBE, and from CC along CFCF, gives

CEEA=[BCP][BAP],AFFB=[CAP][CBP]\frac{CE}{EA} = \frac{[BCP]}{[BAP]}, \qquad \frac{AF}{FB} = \frac{[CAP]}{[CBP]}


In each case the ratio on the side is the ratio of the two triangles around PP that touch that side's ends. The player now shows all three triangles ABPABP, BCPBCP, CAPCAP that tile the big triangle.
ABCPDEF
The other two cevians, frozen

The three triangles around P: CE/EA = [BCP]/[BAP], AF/FB = [CAP]/[CBP].


Why It Holds

Multiply the three results:

BDDC⋅CEEA⋅AFFB=[ABP][ACP]⋅[BCP][BAP]⋅[CAP][CBP]=1\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = \frac{[ABP]}{[ACP]} \cdot \frac{[BCP]}{[BAP]} \cdot \frac{[CAP]}{[CBP]} = 1


Each of the three areas appears exactly once in a numerator and once in a denominator, so the whole product cancels to 11. The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, same-height triangles twice, a subtraction of equal ratios, the same argument twice more, and a multiplication.

The converse follows by a short contradiction: if the product is 11 but the third cevian missed the point where the first two meet, the cevian that does pass through that point would cut the side in the same ratio, forcing it to land at the same point.
ABCPDEF
The conclusion, frozen

Each area once above and once below: the product is 1.


Dragging the Figure

Drag PP anywhere inside the triangle: the three cevians swing, the three feet DD, EE, FF slide along the sides, the three ratios in the readout change, and their product stays at 1.001.00. Drag AA, BB or CC to reshape the triangle. Pull PP towards a side and watch one ratio grow large while another shrinks to compensate.

The player keeps PP strictly inside the triangle, so that all three feet lie on the sides themselves; Ceva's theorem also holds for PP outside, with one or two feet on the extensions of the sides, but then signed ratios are needed, and the player keeps to the inside case. A drag that would carry a vertex out of the frame or flatten the triangle is refused.
ABCPDEF
After dragging P, frozen

P pulled towards B: BD/DC small, AF/FB large, the product still 1.


Consequences

• The triangle centres at once. By the converse, the medians are concurrent (each ratio is 1), the angle bisectors are concurrent at the incentre (the angle bisector theorem makes the product cb⋅ac⋅ba=1\tfrac{c}{b} \cdot \tfrac{a}{c} \cdot \tfrac{b}{a} = 1), and the altitudes are concurrent at the orthocentre (the feet cut the sides in ratios of cosines that cancel). Each of those pages proves its concurrency directly; Ceva proves all three with one calculation.
• Barycentric coordinates. The areas [BCP][BCP], [CAP][CAP], [ABP][ABP], divided by the whole, are the barycentric coordinates of PP; the proof shows that each cevian ratio is a quotient of two of them.
• Menelaus. The companion theorem of Menelaus says the same product is −1-1 (with signed lengths) when DD, EE, FF lie on one line instead of three concurrent lines; it has a similar area proof.
• Gergonne and Nagel points. The cevians to the points where the incircle touches the sides are concurrent (the two tangents from each vertex are equal, so the product is 1); so are the cevians to the points of contact of the excircles.
• Trigonometric form. Replacing each ratio by the ratio of sines of the angles the cevian makes at its vertex gives the trigonometric Ceva, often easier to apply.