Menelaus' Theorem

Given a triangle $ABC$, and a line that crosses $BC$, $AC$ and $AB$ at points $D$, $E$ and $F$ respectively, with $D$, $E$, and $F$ distinct from $A$, $B$ and $C$, then

$\frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA} = -1.$

Proof: What does ratio cries for? Similar triangles!! So let us do some ''ratio chase''.

Important note: There are $2$ different cases to deal with, proving only one of them is not complete. Line $DEF$ may or may not intersect the triangle $ABC$. First, the sign of the left-hand side will be always negative since either all three of the ratios are NEGATIVE, the case where the line $DEF$ do not intersect the triangle or exactly ONE is NEGATIVE and the other TWO are POSITIVE, the case where line $DEF$ crosses TWO sides of the triangle. So to prove both cases, we just need prove the product of magnitudes is $1$.

Case 1: Line DEF intersect the triangle ABC.

Case 2: Line DEF do not intersect the triangle ABC.

Construct perpendiculars from $A$, $B$, and $C$ to the line $DEF$ and let their lengths be $a$, $b$, and $c$ respectively. Then $|\frac{AF}{FB}| = |\frac{a}{b}|$, $|\frac{BD}{DC}| = |\frac{b}{c}|$, and $|\frac{CE}{EA}| = |\frac{c}{a}|$, which we get $|\frac{AF}{FB} \times \frac{BD}{DC} \times \frac{CE}{EA}| = 1$, and we are done. Q.E.D

Here are some problems to pratise with:

1. Let $ABCD$ be a convex quadrilateral such that rays $BA$ and $CD$ intersect at $E$, and rays $AD$ and $BC$ intersect at $F$. Prove that midpoints of $EF$, $AC$, $BD$ are colinear. (This is also known as the Newton's line.)

2. Prove the Desargues' theorem. Let $A_1B_1C_1$ and $A_2B_2C_2$ be two triangles. Then points $P=A_1B_1\cap A_2B_2$, $Q=A_1C_1\cap A_2C_2$ and $R=B_1C_1\cap B_2C_2$ are colinear if and only if lines $A_1A_2$, $B_2B_2$, $C_1C_2$ are concurrent.

This is a strong theorem when dealing with ratios. So try to use it in your future Geometry problems!

(David Altizio) Triangle $ABC$ has $AB=2007$ and $AC=2015$. The incircle $\omega$ of the triangle is tangent to $AC$ and $AB$ at $E$ and $F$ respectively, and $P$ is the intersection point of $EF$ and $BC$. Suppose $B$ is the midpoint of $\overline{CP}$. Compute the length $BC$.

Why would we think to use Menelaus's theorem here?

Well, the point configuration is exactly that of Menelaus! $P, F, E$ are the three collinear points with respect to triangle $ABC$; we are given the ratio $CP/PB$; and we can use some basic properties of the incircle to find some other ratios.

Let $a, b, c$ be the side lengths $BC, CA, AB$ and $s = (a+b+c)/2$ be the semiperimeter, as usual. (We are given $c = 2007, b = 2015,$ and we want to find $a.$) What other lengths can we deduce?

We know that $AF = AE = s-a,$ $FB = s-b,$ and $EC = s-c.$ Therefore, by Menelaus's Theorem, $$\dfrac{CP}{PB} \cdot \dfrac{BF}{FA} \cdot \dfrac{AE}{EC} = 1 \implies 2 \cdot \dfrac{s-b}{s-a} \cdot \dfrac{s-a}{s-c} = 1.$$ This simplifies to $2(s-b) = s-c.$ Then $2s - 2b = s - c,$ so $s = 2b - c.$ Hence $(a+b+c)/2 = 2b - c,$ and solving for $a,$ we get $a = 3b - 3c.$ With $c = 2007$ and $b = 2015,$ we find that $a = 3(2015-2007) = 24.$ $\square$