**Ceva's Theorem**and

**Menelaus' Theorem**. These two theorems are very useful in plane geometry because we often use them to prove that a certain number of

*points lie on a straight line*and a certain number of

*lines intersect at a single point*. Both of the theorems will be proved based on a common simple principle. We also generalize the theorems for

*arbitrary polygons*.

Let us state the theorems.

Ceva's Theorem:Given a triangle $ABC$ and three points $A'$, $B'$, $C'$ lying on the threelines$BC$, $CA$, $AB$, respectively. Then the three lines $AA'$, $BB'$, $CC'$ areconcurrentif and only if $$\frac{\vec{A'B}}{\vec{A'C}} \times \frac{\vec{B'C}}{\vec{B'A}} \times \frac{\vec{C'A}}{\vec{C'B}} = -1.$$

Menelaus' Theorem:Given a triangle $ABC$ and three points $A'$, $B'$, $C'$ lying on the threelines$BC$, $CA$, $AB$, respectively. Then the three points $A'$, $B'$, $C'$ arecollinearif and only if $$\frac{\vec{A'B}}{\vec{A'C}} \times \frac{\vec{B'C}}{\vec{B'A}} \times \frac{\vec{C'A}}{\vec{C'B}} = 1.$$

Signed ratio of line segments

First, let us explain a little bit about the notation that we use in the theorems. The notation $\frac{\vec{A'B}}{\vec{A'C}}$ is called a

**signed ratio**. In the figure below, we can see that the

*normal (unsigned) ratio*$\frac{UX}{UY} = 2$, but the

*signed ratio*is $\frac{\vec{UX}}{\vec{UY}} = -2$. This is because $\vec{UX}$ and $\vec{UY}$ point to

*opposite directions*.

The

*normal ratio*is always a positive number but the

*signed ratio*can be negative or positive. The signed ratio $\frac{\vec{UX}}{\vec{UY}}$ is

*positive*when $\vec{UX}$ and $\vec{UY}$ point to the

*same direction*, and it is

*negative*when $\vec{UX}$ and $\vec{UY}$ points to

*opposite directions*.

*So why do we need the signed ratio?*The trouble with our normal ratio is that it

*cannot*uniquely identify a point on a line in relation to two other reference points, but the signed ratio can.

Let us consider an example. Suppose on the line $XY$, we want to specify a point $Z$ so that $\frac{ZX}{ZY} = 2$.

In the above figure, we can see that using our usual unsigned ratio, there are two points, $U$ and $V$, satisfying this condition since $$\frac{UX}{UY} = \frac{VX}{VY} = 2.$$

But if we consider

*signed ratio*, then there is a

*unique point*, which is $V$ satisfying $\frac{\vec{VX}}{\vec{VY}} = 2$.

The point $U$ no longer satisfies because $\frac{\vec{UX}}{\vec{UY}} = -2 \neq 2$.

Signed area

Similar to signed ratio, we have the concept of

**signed area**. The normal (unsigned) area of a polygon is always a positive number but a signed area can be positive or negative depending on the

*orientation*of the vertices of the polygon.

If we draw a coordinate system $0xy$ on the plane then any point $A$ on the plane will have an associated coordinate $(A_x, A_y)$. We define $$[A,B] = A_x B_y - A_y B_x$$ and define the

*signed area*of a triangle $ABC$ to be $$\overline{s}(ABC) = \frac{1}{2}([A,B] + [B,C] + [C,A]).$$

In the above figure, we have the coordinates $A(-5,2)$, $B(-3,7)$, $C(3,2)$. So we can calculate $$[A,B] = A_x B_y - A_y B_x = (-5) (7) - (2) (-3) = -29,$$ $$[B,C] = B_x C_y - B_y C_x = (-3) (2) - (7) (3) = -27,$$ $$[C,A] = C_x A_y - C_y A_x = (3) (2) - (2) (-5) = 16.$$ Thus, $$\overline{s}(ABC)= \frac{1}{2}([A,B] + [B,C] + [C,A])= \frac{1}{2}(-29-27+16)=-20.$$

On the other hand, $[A,C]=-16$, $[C,B]=27$, $[B,A]=29$ and $\overline{s}(ACB)= 20$.

A theorem on ratio of signed areas

Let us now state a simple theorem concerning about the ratio of the signed areas of two triangles that share a common edge. We will use this theorem later to prove the Ceva's theorem and the Menelaus' theorem.

A Theorem on Ratio of Signed Areas.Given two triangles $ABU$ and $ABV$ that share a common edge $AB$. The line connecting the two vertices $UV$ intersects with the line $AB$ at a point $T$. Then it must hold that $$\frac{\overline{s}(ABU)}{\overline{s}(ABV)} = \frac{\vec{TU}}{\vec{TV}}.$$

This theorem is quite obvious if we forget about the sign and consider the usual unsigned ratio. This is because if we draw the two altitudes $UU'$ and $VV'$ perpendicular to the side $AB$ then $$\frac{s(ABU)}{s(ABV)} = \frac{AB \times UU' / 2}{AB \times VV' /2} = \frac{UU'}{VV'} = \frac{TU}{TV}.$$ We will give a proof of this theorem for the signed ratio in the next post.

Proving Ceva's Theorem and Menelaus' Theorem

Let us now use the above theorem to prove Ceva's Theorem and Menelaus' Theorem.

**Proof of Ceva's Theorem.**Suppose the three lines $AA'$, $BB'$, $CC'$ are concurrent at a point $I$. Then by the theorem on the ratio of signed areas, we have $$\frac{\vec{A'B}}{\vec{A'C}} = \frac{\overline{s}(IAB)}{\overline{s}(IAC)}, ~~~~\frac{\vec{B'C}}{\vec{B'A}} = \frac{\overline{s}(IBC)}{\overline{s}(IBA)}, ~~~~\frac{\vec{C'A}}{\vec{C'B}} = \frac{\overline{s}(ICA)}{\overline{s}(ICB)}.$$

Thus, $$\frac{\vec{A'B}}{\vec{A'C}} \times \frac{\vec{B'C}}{\vec{B'A}} \times \frac{\vec{C'A}}{\vec{C'B}} = \frac{\overline{s}(IAB)}{\overline{s}(IAC)} \times \frac{\overline{s}(IBC)}{\overline{s}(IBA)} \times \frac{\overline{s}(ICA)}{\overline{s}(ICB)}$$ $$= \left( - \frac{\overline{s}(IAB)}{\overline{s}(ICA)} \right) \times \left( - \frac{\overline{s}(IBC)}{\overline{s}(IAB)} \right) \times \left( - \frac{\overline{s}(ICA)}{\overline{s}(IBC)} \right) = -1.$$

In the reverse direction, if $$\frac{\vec{A'B}}{\vec{A'C}} \times \frac{\vec{B'C}}{\vec{B'A}} \times \frac{\vec{C'A}}{\vec{C'B}} = -1,$$ we need to prove that the three lines $AA'$, $BB'$, $CC'$ are concurrent. In this case, let $AA'$ and $BB'$ intersect at a point $I$, let $CI$ and $AB$ intersect at $C''$, we need to show that $C'' = C'$. Indeed, since $AA'$, $BB'$, $CC''$ are concurrent, by what we have just proved, we have $$\frac{\vec{A'B}}{\vec{A'C}} \times \frac{\vec{B'C}}{\vec{B'A}} \times \frac{\vec{C''A}}{\vec{C''B}} = -1,$$ thus, $$\frac{\vec{C''A}}{\vec{C''B}} = \frac{\vec{C'A}}{\vec{C'B}}.$$ Since the signed ratio determines a point uniquely on the line $AB$, we obtain $C' = C''$.

So we have proved the Ceva's Theorem.

**Proof of Menelaus' Theorem.**Suppose that the three points $A'$, $B'$, $C'$ are collinear. Take two arbitrary points $I$ and $J$ on the line $A'B'C'$, then by the theorem on the ratio of signed areas, we have $$\frac{\vec{A'B}}{\vec{A'C}} = \frac{\overline{s}(IJB)}{\overline{s}(IJC)}, ~~~~\frac{\vec{B'C}}{\vec{B'A}} = \frac{\overline{s}(IJC)}{\overline{s}(IJA)}, ~~~~\frac{\vec{C'A}}{\vec{C'B}} = \frac{\overline{s}(IJA)}{\overline{s}(IJB)}.$$

Thus, $$\frac{\vec{A'B}}{\vec{A'C}} \times \frac{\vec{B'C}}{\vec{B'A}} \times \frac{\vec{C'A}}{\vec{C'B}} = \frac{\overline{s}(IJB)}{\overline{s}(IJC)} \times \frac{\overline{s}(IJC)}{\overline{s}(IJA)} \times \frac{\overline{s}(IJA)}{\overline{s}(IJB)} = 1.$$

The reverse direction of Menelaus' Theorem is proved similar to that of Ceva's Theorem.

So we have finally proved Ceva's Theorem and Menelaus' Theorem. This proof also works if we only consider the usual unsigned ratio. We remark that for Menelaus' theorem, instead of using the ratio of the areas of the triangles, the same result can be obtained if we use the

*ratio of the altitudes*drawing from $A$, $B$, $C$ perpendicular to the line $A'B'C'$.

Generalization of Ceva's Theorem and Menelaus' Theorem

The above proof of Ceva's theorem and Menelaus' theorem motivates us to generalize the theorems for arbitrary polygons.

Here are Ceva's theorem and Menelaus' theorem for pentagon.

**Ceva's theorem for pentagon.**Given a pentagon $A_1 A_2 A_3 A_4 A_5$ and five points $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ lying on the lines $A_5 A_2$, $A_1 A_3$, $A_2 A_4$, $A_3 A_5$, $A_4 A_1$, respectively. If the lines $A_1 B_1$, $A_2 B_2$, $A_3 B_3$, $A_4 B_4$, $A_5 B_5$ are concurrent then $$\frac{\vec{B_1 A_5}}{\vec{B_1 A_2}} \times \frac{\vec{B_2 A_1}}{\vec{B_2 A_3}} \times \frac{\vec{B_3 A_2}}{\vec{B_3 A_4}} \times \frac{\vec{B_4 A_3}}{\vec{B_4 A_5}} \times \frac{\vec{B_5 A_4}}{\vec{B_5 A_1}}= -1.$$

**Proof.**If the five lines $A_1 B_1$, $A_2 B_2$, $A_3 B_3$, $A_4 B_4$, $A_5 B_5$ are concurrent at a point $I$ then $$\frac{\vec{B_1 A_5}}{\vec{B_1 A_2}} \times \frac{\vec{B_2 A_1}}{\vec{B_2 A_3}} \times \frac{\vec{B_3 A_2}}{\vec{B_3 A_4}} \times \frac{\vec{B_4 A_3}}{\vec{B_4 A_5}} \times \frac{\vec{B_5 A_4}}{\vec{B_5 A_1}}$$ $$= \frac{\overline{s}(I A_1 A_5)}{\overline{s}(I A_1 A_2)} \times \frac{\overline{s}(I A_2 A_1)}{\overline{s}(I A_2 A_3)} \times \frac{\overline{s}(I A_3 A_2)}{\overline{s}(I A_3 A_4)} \times \frac{\overline{s}(I A_4 A_3)}{\overline{s}(I A_4 A_5)} \times \frac{\overline{s}(I A_5 A_4)}{\overline{s}(I A_5 A_1)}$$ $$= \left( - \frac{\overline{s}(I A_5 A_1)}{\overline{s}(I A_1 A_2)} \right) \left( - \frac{\overline{s}(I A_1 A_2)}{\overline{s}(I A_2 A_3)} \right) \left( - \frac{\overline{s}(I A_2 A_3)}{\overline{s}(I A_3 A_4)} \right) \left( - \frac{\overline{s}(I A_3 A_4)}{\overline{s}(I A_4 A_5)} \right) \left( - \frac{\overline{s}(I A_4 A_5)}{\overline{s}(I A_5 A_1)} \right) = -1.$$

**Menelaus' theorem for pentagon.**Given a pentagon $A_1 A_2 A_3 A_4 A_5$ and five points $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ lying on the lines $A_1 A_2$, $A_2 A_3$, $A_3 A_4$, $A_4 A_5$, $A_5 A_1$, respectively. If the points $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ are collinear then $$\frac{\vec{B_1 A_1}}{\vec{B_1 A_2}} \times \frac{\vec{B_2 A_2}}{\vec{B_2 A_3}} \times \frac{\vec{B_3 A_3}}{\vec{B_3 A_4}} \times \frac{\vec{B_4 A_4}}{\vec{B_4 A_5}} \times \frac{\vec{B_5 A_5}}{\vec{B_5 A_1}}= 1.$$

**Proof.**If the five points $B_1$, $B_2$, $B_3$, $B_4$, $B_5$ lie on the same straight line then take two arbitrary points $I$, $J$ on this line, we have $$\frac{\vec{B_1 A_1}}{\vec{B_1 A_2}} \times \frac{\vec{B_2 A_2}}{\vec{B_2 A_3}} \times \frac{\vec{B_3 A_3}}{\vec{B_3 A_4}} \times \frac{\vec{B_4 A_4}}{\vec{B_4 A_5}} \times \frac{\vec{B_5 A_5}}{\vec{B_5 A_1}}$$ $$= \frac{\overline{s}(I J A_1)}{\overline{s}(I J A_2)} \times \frac{\overline{s}(I J A_2)}{\overline{s}(I J A_3)} \times \frac{\overline{s}(I J A_3)}{\overline{s}(I J A_4)} \times \frac{\overline{s}(I J A_4)}{\overline{s}(I J A_5)} \times \frac{\overline{s}(I J A_5)}{\overline{s}(I J A_1)} = 1.$$

Here are Ceva's Theorem and Menelaus' Theorem for general polygons.

Ceva's Theorem for polygon.Given a polygon $A_1 A_2 \dots A_n$ and $n$ points $B_1$, ..., $B_n$, where each point $B_i$ lies on the line $A_{i-1} A_{i+1}$. If $n$ lines $A_1 B_1$, $A_2 B_2$, ..., $A_n B_n$ are concurrent then $$\prod_{i=1}^{n} \frac{\vec{B_i A_{i-1}}}{\vec{B_i A_{i+1}}} = (-1)^n.$$

Menelaus' Theorem for polygon.Given a polygon $A_1 A_2 \dots A_n$ and $n$ points $B_1$, ..., $B_n$, where each point $B_i$ lies on the line $A_i A_{i+1}$. If the points $B_1$, $B_2$, ..., $B_n$ are collinear then $$\prod_{i=1}^{n} \frac{\vec{B_i A_i}}{\vec{B_i A_{i+1}}} = 1.$$

Today we have learned about Ceva's Theorem and Menelaus' Theorem. Both of these theorems are proved by using a simple

*theorem on the ratio of triangle's areas*. This proof is so simple, yet it helps us to generalize the theorems for arbitrary polygons.

Let us stop here for now, hope to see you again in the next post.

*Homework.*

1. In the following figure, prove that $$\frac{UB}{UC} = \frac{VB}{VC}.$$

2. Given a triangle $ABC$ with the lengths of the sides $AB = c$, $BC = a$, $CA = b$. Suppose that the incircle touches the sides of the triangle at $A'$, $B'$, $C'$. Determine the lengths of $AB'$, $AC'$, $BA'$, $BC'$, $CA'$, $CB'$ in terms of $a$, $b$, $c$. Prove that the three lines $AA'$, $BB'$, $CC'$ are concurrent.

3. Generalize Menelaus' Theorem for points in 3D space. Below is a version of Menelaus' Theorem for tetrahedron.

Given a tetrahedron $ABCD$. A plane intersects the four lines $AB$, $BC$, $CD$, $DA$ at the points $X$, $Y$, $Z$, $T$, respectively. Prove that $$\frac{\vec{XA}}{\vec{XB}} \times \frac{\vec{YB}}{\vec{YC}} \times \frac{\vec{ZC}}{\vec{ZD}} \times \frac{\vec{TD}}{\vec{TA}} = 1.$$

5. Take some examples of three points $A$, $B$, $C$ lying on a same straight line on a coordinate system $0xy$ and calculate the signed area $\overline{s}(ABC)$.

6. Prove that $[A,B] = -[B,A]$, $[A,A] = 0$ and $\overline{s}(ABC) = -\overline{s}(ACB)$.

7. Suppose $O$ is the center of a coordinate system $0xy$. Prove that $$\overline{s}(OAB) = \frac{1}{2} [A,B], ~~~~\overline{s}(OBC) = \frac{1}{2} [B,C], ~~~~\overline{s}(OCA) = \frac{1}{2} [C,A],$$

it follows that $$\overline{s}(ABC) = \overline{s}(OAB) + \overline{s}(OBC) + \overline{s}(OCA).$$

Using the above identity to prove that for any point $M$, $$\overline{s}(ABC) = \overline{s}(MAB) + \overline{s}(MBC) + \overline{s}(MCA)$$

8. Take some examples of four points $A$, $B$, $C$, $D$ on a coordinate system $0xy$ and then calculate the signed area $$\overline{s}(ABCD) = \frac{1}{2}([A,B] + [B,C] + [C,D] + [D,A]).$$

Check if the normal unsigned area $s(ABCD)$ is agreeable to the signed area $\overline{s}(ABCD)$.

Generalize the concept of signed area for polygons.

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