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Cevian is called every segment that connects a vertice of a triangle with the opposite side. Three or more lines are concurrent when all pass through a common point, so we are talking about concurence.

Historical Facts

This theorem was published from Giovanni Ceva (in honour of whom the segments before are named ) in 1678.

Theorem

In any triangle $ABC$ the cevians $AD, BE, CF$ are concurrent if and only if

$\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1$ (simple form)

or

$\frac{sinBAD}{sinDAC}\cdot\frac{sinABE}{sinEBC}\cdot\frac{sinBCF}{sinFCA}=1$ (trigonometric form)

Proofs

Proof 1

First we prove that we have concurrency if and only if $\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1$. Assume that AD, BE, CF meet at point P. We extend BE and CF until they meet the parallel to BC from A at G and H respectively.

Then, we have the relations

$\frac{AF}{FB}= \frac{AH}{BC}$ (1) $\frac{CE}{EA}= \frac{BC}{AG}$ (2) and $\frac{BD}{DC}= \frac{BD}{DP} \cdot \frac{DP}{DC}= \frac{AG}{PA} \cdot \frac{PA}{AH}$ (3) Multipling relations 1, 2 and 3 we get the desired.

Now suppose that $\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1$. We have to prove that AD, BE, CF are concurrent. Let BE and CF meet at P and suppose AP cuts BC at D’.

We get by what we have proved so far that  $\frac{BD'}{D'C}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1$. Dividing the last two relations we get that $\frac{BD}{DC} = \frac{BD'}{D'C}$ and since D and D’ both lie in segment BC, we get $D \equiv D'$, QED

Proof 2 (Using areas)

We use the fact that the areas of triangles with equal altitudes are proportional to the bases of the triangles. Let AD, BE, CF concur at P. We have :

$\frac{BD}{DC}= \frac{(BDA)}{(CDA)}= \frac{(BDP)}{(DCP)}$ $\Leftrightarrow \frac{BD}{DC}= \frac{(BDA)-(BDP)}{(CDA)-(CDP)}= \frac{(APB)}{(APC)}$. Similarly we get $\frac{CE}{CA}= \frac{(BCP)}{(BAP)}$ and $\frac{AF}{FB}= \frac{(CAP)}{(CBP)}$. Multipling these relations we get the desired $\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1$.

For the converse we work like the first proof to get $D \equiv D'$.

Proof of the trigonometric form of Ceva’s theorem

If AD, BE, CF are concurrent, say at P, by the Law of Sines at triangles ABP, BCP, CPA we get $\frac{sinBAP}{sinABP}=\frac{BP}{AP}$ (1) $\frac{sinCBP}{sinBCP}= \frac{CP}{BP}$ (2) $\frac{sinACP}{sinCAP}= \frac{AP}{CP}$ (3) and multipling these we get the desired.

Finally, we have to show that if $\frac{sinBAD}{sinDAC}\cdot\frac{sinABE}{sinEAC}\cdot\frac{sinBCF}{sinFCA}=1$ then AD, BE, CF concur.

By the Law of Sines at triangles ABD and ACD we get $\frac{sinBAD}{sinBDA}= \frac{BD}{AB}$ (1) and $\frac{sinCAD}{sinCDA}= \frac{CD}{AC}$ (2). But $sinBDA=sinCDA$ since they are complementa angles. So, dividing relations 1 and 2,  $\frac{sinBAD}{sinCDA}= \frac{BD}{DC} \cdot \frac{AB}{CA}$ . Similarly, $\frac{sinABE}{sinCBE}= \frac{CE}{EA} \cdot \frac{BC}{BA}$ and $\frac{sinBCF}{sinACF}= \frac{AF}{FB} \cdot \frac{CA}{CB}$. Multipling the last three relations and using the hypothesis we get $\frac{BD}{DC}\cdot\frac{CE}{EA}\cdot\frac{AF}{FB}=1$ which is enough for what we need since we have prove it earlier.

Notes

– The points D, E, F can be on the extensions of the sides of the triangle

Results

• The medians of a triangle concur at the centroid. Simply aplly Ceva’s theorem using $BM=MC, CN=NA, AL=LB$ where $M, N, L$ are the midpoints of $BC, CA, AB$ respectively.
• The internal angle bisectors at the incenter. By the intarnal angle bisector theorem, if  $AD$ the bisector of $\widehat{BCA}$ then $\frac{BD}{DC}=\frac{AB}{AC}$ etc. Multiply the 3 equations and QED.
• The three altitudes of every triangle concur at the orthocenter of the triangle. If $D, E, F$ are the projections of $A, B, C$ at the opposite sides, use the trigonometric form of Ceva with $\widehat{BAD}=\widehat{BCF}, \widehat{ABE}=\widehat{ACF}$ and $\widehat{CAD}=\widehat{CBE}$
• The cevians joining the points of tangency of the incircle with the sides of a triangle concur at the Gergonne point of the triangle. If $D, E, F$ the desired points, then $AF=AE, BF=BD, CD=CE$ so by Ceva’s simple form the conclusion is obvious.
• Similarly the cevians that connect the points of tangency of the 3 excircles with the corresponding sides of a tringle also concur at Nagel point. (The proof comes soon)

Exercises

1. Let the ceavians AD, BE, CF be concurrent in a triangle ABC. Prove that FE is parallel to BC if and only if AD is the bisector of BC
2. In a triangle ABC the median AM meets the internal bisector BN at P. Let Q be the intersection of AB and CP. Prove that BNQ is an isosceles triangle.
3. Let ABC be a triangle and let D, E, F be the points on sides BC, CA, AB respectively such that the cevians AD, BE, CF are concurrent. Let M, N, P be points on EF, FD, DE respectively. Prove that lines AM, BN, CP concur if and only if lines DM, EN, FP concur.
4. (*)In a triangle ABC points D, E, F are on sides BC, CA, AB so that cevians AD, BE, CF concur. Let the parallels to the line AB from points E and Q meet the lines DF and EF at Q and R respectively. Prove that lines CF, DE, QR are concurrent.

$latex \frac{BD}{DC}= \frac{(BDA)}{(CDA)}= \frac{(BDP)}{(DCP)}=$ . Similarly we get $\frac{CE}{CA}= \frac{(BCP)}{(BAP)}$ and $latex \frac{AF}{FB}= \frac{(CAP)}{( C BP)}$. Multipling these relations we get the desired $\frac{AF}{FB}\cdot\frac{BD}{DC}\cdot\frac{CE}{EA}=1$

For the converse we work like the first prove to get $D \equiv D'$

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