Ceva's Theorem

Ceva's theorem. Consider a triangle $ABC$ with three cevians drawn as follows. Then, the cevians $\overline{AD}$, $\overline{BE}$, and $\overline{CF}$ concur if and only if $$\frac {AF}{FB} \cdot \frac {BD}{DC} \cdot \frac {CE}{EA} = 1.$$

Proof. (AoPSWiki) We will use the notation $[ABC]$ to denote the area of a triangle with vertices $A,B,C$.

First, suppose $AD, BE, CF$ meet at a point $X$. We note that triangles $ABD, ADC$ have the same altitude to line $BC$, but bases $BD$ and $DC$. It follows that $$\frac {BD}{DC} = \frac{[ABD]}{[ADC]}.$$ The same is true for triangles $XBD, XDC$, so $$\frac{BD}{DC} = \frac{[ABD]}{[ADC]} = \frac{[XBD]}{[XDC]} = \frac{[ABD]- [XBD]}{[ADC]-[XDC]} = \frac{[ABX]}{[AXC]}.$$ Similarly, $\frac{CE}{EA} = \frac{[BCX]}{[BXA]}$ and $\frac{AF}{FB} = \frac{[CAX]}{[CXB]}$, so $$\frac{BD}{DC} \cdot \frac{CE}{EA} \cdot \frac{AF}{FB} = \frac{[ABX]}{[AXC]} \cdot \frac{[BCX]}{[BXA]} \cdot \frac{[CAX]}{[CXB]} = 1.$$

Now, suppose $D,E,F$ satisfy Ceva's criterion, and suppose $AD, BE$ intersect at $X$. Suppose the line $CX$ intersects line $AB$ at $F'$. We have proven that $F'$ must satisfy Ceva's criterion. This means that $$\frac{AF'}{F'B} = \frac{AF}{FB},$$ so $F' = F$ and line CF concurs with $AD$ and $BE$. $\blacksquare$

Note: there is also a proof with barycentric coordinates, which you will find out about when you reach that section!

Ceva's Theorem

Ceva's theorem is a particularly useful problem solving tool in geometry problems involving triangles, and in some cases even polygons that could be reduced to triangles.

A Cevian is any line that connects a vertex of a triangle to a point on the opposite side. Therefore, in the above triangle, $CK, BL$ and $AM$ are Cevians.

Ceva's theorem captures how well-behaved Cevians are when they are concurrent. And the above image illustrates 3 concurrent Cevians.

The theorem's statement (with reference to the above triangle's labels) goes as follows:

$\frac{CM\cdot BK \cdot AL}{BM \cdot AK \cdot LC}=1$ if an only if the Cevians $AM$, $BL$ and $CK$ are concurrent.

Try proving this yourself before you proceed! Then compare the proof given here with yours.

Notice the triangles that the Cevians create? Try playing with area ratios.

Here is a simple proof:

If the Cevians are concurrent, then $$\frac{BK}{AK}=\frac{\operatorname*{ar}(BKC)}{\operatorname*{ar}(AKC)}$$

Also, $\frac{BK}{AK} = \frac{\operatorname*{ar}(OBK)}{\operatorname*{ar}(OAK)}$.

As a result, $\frac{BK}{AK}$ $= \frac{\operatorname*{ar}(BKC)-\operatorname*{ar}(OBK)}{\operatorname*{ar}(AKC)-\operatorname*{ar}(OAK)}$

$=\frac{\operatorname*{ar}(BOC)}{\operatorname*{ar}(AOC)}$

Now, applying a similar computation on other sides yields:

$$\frac{CM}{BM}=\frac{\operatorname*{ar}(AOC)}{\operatorname*{ar}(BOA)}$$ $$\frac{AL}{CL}=\frac{\operatorname*{ar}(BOA)}{\operatorname*{ar}(BOC)}$$

Combining the three equalities by multiplying them with eachother, we get:

$$$\frac{CM\cdot BK \cdot AL}{BM \cdot AK \cdot LC}=\frac{\operatorname*{ar}(BOC)\operatorname*{ar}(BOA)\operatorname*{ar}(AOC)}{\operatorname*{ar}(AOC)\operatorname*{ar}(BOC)\operatorname*{ar}(BOA)}$

By cancellation, the right hand side of the above equation becomes $1$.

And if you proved it using a different idea from the one used here, do let us know in the comments!

As a simple exercise try showing that medians of a triangle are concurrent using Ceva's theorem.

Proving Stewart's theorem

Try using what you know about Cevians to prove Stewart's theorem.

Let $CK$ be a Cevian in the triangle $ABC$. Try to show the following. $$AB(CK^2 + BK\cdot AK) = BK\cdot AC^2 + AK\cdot BC^2$$

Compare your proof with the one given here.

Let $\theta = \operatorname*{angle }CKB$

By applying Cosine rule, we get

$$BK^2 + CK^2 - 2BK\cdot CK \cos \theta=BC^2$$ $$CK^2 + AK^2 - 2CK\cdot AK\cdot (180-\theta) = CK^2 + AK^2 + 2CK\cdot AK\cdot (\theta)$$

Now, multiply the first equation by $AK$ and the second equation by $BK$ and add them.

We get the following soup when we do that.

$$BK^2AK+CK^2AK+CK^2BK+AK^2BK=BC^2AK+AC^2BK$$

And this evaluates to $$AK\cdot BK(BK+AK)+CK^2(AK+BK)=BC^2AK+AC^2BK$$ $$AB(AK\cdot BK+CK^2)=BC^2AK+AC^2BK$$

Let us know in comments if you thought of a different proof.