Ceva's Theorem

Ceva's Theorem is a criterion for the concurrence of cevians in a triangle.

Statement

Let be a triangle, and let be points on lines , respectively. Lines are concurrent if and only if where lengths are directed. This also works for the reciprocal of each of the ratios, as the reciprocal of is .

(Note that the cevians do not necessarily lie within the triangle, although they do in this diagram.)

The proof using Routh's Theorem is extremely trivial, so we will not include it.

Proof

We will use the notation to denote the area of a triangle with vertices .

First, suppose meet at a point . We note that triangles have the same altitude to line , but bases and . It follows that . The same is true for triangles , so Similarly, and , so Now, suppose satisfy Ceva's criterion, and suppose intersect at . Suppose the line intersects line at . We have proven that must satisfy Ceva's criterion. This means that $$ (Error compiling LaTeX. Unknown error_msg)\frac{AF'}{F'B} = \frac{AF}{FB} CFADBE\square$== Proof by [[Barycentric coordinates]] ==

Since$ (Error compiling LaTeX. Unknown error_msg)D\in BC(0,d,1-d)ADz=\frac{1-d}{d}y$.

Similarly, since$ (Error compiling LaTeX. Unknown error_msg)E=(1-e,0,e)F=(f,1-f,0)BECFx=\frac{1-e}{e}zy=\frac{1-f}{f}x$Multiplying the three together yields the solution to the equation: <cmath>xyz=\frac{1-e}{e}\cdot{z}\cdot\frac{1-f}{f}\cdot{x}\cdot\frac{1-d}{d}y</cmath>

Dividing by$ (Error compiling LaTeX. Unknown error_msg)xyz$yields: <cmath>1=\frac{1-e}{e}\cdot\frac{1-f}{f}\cdot\frac{1-d}{d}</cmath>, which is equivalent to Ceva's Theorem.

<cmath>Q.E.D.</cmath>

== Trigonometric Form ==

The [[trigonometry|trigonometric]] form of Ceva's Theorem states that cevians$ (Error compiling LaTeX. Unknown error_msg)AD,BE,CF$concur if and only if <cmath>\frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = 1</cmath>

=== Proof ===

First, suppose$ (Error compiling LaTeX. Unknown error_msg)AD, BE, CF X $. We note that <cmath>\frac{[BAX]}{[XAC]} = \frac{ \frac{1}{2}AB \cdot AX \cdot \sin BAX}{ \frac{1}{2}AX \cdot AC \cdot \sin XAC} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC}</cmath> and similarly, <cmath>\frac{[CBX]}{[XBA]} = \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} ;\; \frac{[ACX]}{[XCB]} = \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB}</cmath> It follows that <cmath> \frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} \cdot \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} \cdot \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB} </cmath> <br> <cmath> \qquad = \frac{[BAX]}{[XAC]} \cdot \frac{[CBX]}{[XBA]} \cdot \frac{[ACX]}{[XCB]} = 1</cmath>

Here, the sign is irrelevant, as we may interpret the sines of [[directed angles]] mod$ (Error compiling LaTeX. Unknown error_msg)\pi $to be either positive or negative.

The converse follows by an argument almost identical to that used for the first form of Ceva's theorem.$ (Error compiling LaTeX. Unknown error_msg)\square$== Problems ==

=== Introductory ===

• Suppose$ (Error compiling LaTeX. Unknown error_msg)AB, ACBC13, 1415\frac{AF}{FB} = \frac{2}{5}\frac{CE}{EA} = \frac{5}{8}BDDC$. ([[Ceva's Theorem/Problems|Source]])

=== Intermediate ===

• In$ (Error compiling LaTeX. Unknown error_msg)\Delta ABC, AD, BE, CFP, Q, REF, FD, DEDP, EQ, FRAP, BQ, CR$are concurrent.

• Let$ (Error compiling LaTeX. Unknown error_msg)MABABCDEBCCADEABPAMEMCPXDPCMYX,Y,B$ are collinear. (Source)

See Also

• Stewart's Theorem
• Menelaus' Theorem
• Routh's Theorem