# Intercept theorem

Not to be confused with Thales' theorem.

The intercept theorem, also known as Thales' theorem (not to be confused with another theorem with that name), is an important theorem in elementary geometry about the ratios of various line segments that are created if two intersecting lines are intercepted by a pair of parallels. It is equivalent to the theorem about ratios in similar triangles. Traditionally it is attributed to Greek mathematician Thales.[1]

## Formulation

Suppose S is the intersection point of two lines and A, B are the intersections of the first line with the two parallels, such that B is further away from S than A, and similarly C, D are the intersections of the second line with the two parallels such that D is further away from S than C.

1. The ratios of any two segments on the first line equals the ratios of the according segments on the second line: ${\displaystyle |SA|:|AB|=|SC|:|CD|}$, ${\displaystyle |SB|:|AB|=|SD|:|CD|}$, ${\displaystyle |SA|:|SB|=|SC|:|SD|}$
2. The ratio of the two segments on the same line starting at S equals the ratio of the segments on the parallels: ${\displaystyle |SA|:|SB|=|SC|:|SD|=|AC|:|BD|}$
The converse of the first statement is true as well, i.e. if the two intersecting lines are intercepted by two arbitrary lines and ${\displaystyle |SA|:|AB|=|SC|:|CD|}$ holds then the two intercepting lines are parallel. However the converse of the second statement is not true.
3. If you have more than two lines intersecting in S, then ratio of the two segments on a parallel equals the ratio of the according segments on the other parallel: ${\displaystyle |AF|:|BE|=|FC|:|ED|}$ , ${\displaystyle |AF|:|FC|=|BE|:|ED|}$
An example for the case of three lines is given the second graphic below.

The first intercept theorem shows the ratios of the sections from the lines, the second the ratios of the sections from the lines as well as the sections from the parallels, finally the third shows the ratios of the sections from the parallels.

## Related concepts

### Similarity and similar triangles

Arranging two similar triangles, so that the intercept theorem can be applied

The intercept theorem is closely related to similarity. It is equivalent to the concept of similar triangles, i.e. it can be used to prove the properties of similar triangles and similar triangles can be used to prove the intercept theorem. By matching identical angles you can always place two similar triangles in one another so that you get the configuration in which the intercept theorem applies; and conversely the intercept theorem configuration always contains two similar triangles.

### Scalar multiplication in vector spaces

In a normed vector space, the axioms concerning the scalar multiplication (in particular ${\displaystyle \lambda \cdot ({\vec {a}}+{\vec {b}})=\lambda \cdot {\vec {a}}+\lambda \cdot {\vec {b}}}$ and ${\displaystyle \|\lambda {\vec {a}}\|=|\lambda |\cdot \ \|{\vec {a}}\|}$) are assuring that the intercept theorem holds. One has ${\displaystyle {\frac {\|\lambda \cdot {\vec {a}}\|}{\|{\vec {a}}\|}}={\frac {\|\lambda \cdot {\vec {b}}\|}{\|{\vec {b}}\|}}={\frac {\|\lambda \cdot ({\vec {a}}+{\vec {b}})\|}{\|{\vec {a}}+{\vec {b}}\|}}=|\lambda |}$

## Applications

### Algebraic formulation of compass and ruler constructions

There are three famous problems in elementary geometry which were posed by the Greeks in terms of Compass and straightedge constructions:[2]

Their solution took more than 2000 years until all three of them finally were settled in the 19th century using algebraic methods that had become available during that period of time. In order to reformulate them in algebraic terms using field extensions, one needs to match field operations with compass and straightedge constructions. In particular it is important to assure that for two given line segments, a new line segment can be constructed such that its length equals the product of lengths of the other two. Similarly one needs to be able to construct, for a line segment of length ${\displaystyle d}$, a new line segment of length ${\displaystyle d^{-1}}$. The intercept theorem can be used to show that in both cases such a construction is possible.

 Construction of a product Construction of an inverse

### The construction of a decimal number

A practical example of the third intercept theorem according to formulation point 3. in combination with number line.

 The basic construction consists of a horizontal number line ${\displaystyle s_{1}}$ with the dividing point ${\displaystyle 1,}$ two vertical parallel number line ${\displaystyle s_{2},s_{3}}$ both with ten division points and a diagonal that the vertex generated from the point ${\displaystyle 10}$ of ${\displaystyle s_{3}}$ through the point ${\displaystyle 1}$ of ${\displaystyle s_{2}}$ ${\displaystyle K_{1}}$ on ${\displaystyle s_{1}.}$ 1) As an example, the construction of the decimal number ${\displaystyle 8.6}$ Connect the dividing point ${\displaystyle 6}$ of ${\displaystyle s_{3}}$ with ${\displaystyle K_{1},}$ there arises the value ${\displaystyle 0.6}$ on ${\displaystyle s_{2}.}$ Transfer (added) the value ${\displaystyle 0.6}$ from the dividing point ${\displaystyle 8}$ of ${\displaystyle s_{2},}$ there arises the value ${\displaystyle 8.6.}$ Draw a circular arc from the value ${\displaystyle 8.6}$ to ${\displaystyle s_{1}}$ on, it gives the value ${\displaystyle 8.6}$ on ${\displaystyle s_{1}.}$ Thus, the decimal number ${\displaystyle 8.6}$ is constructed. 1) For the representation of a decimal number with decimal places > 1, the construction of a vertex is K2 (analogous K1) across from K1 advantageous. See also Commons: Decimal number 8.639

### Dividing a line segment in a given ratio

 To divide an arbitrary line segment ${\displaystyle {\overline {AB}}}$ in a ${\displaystyle m:n}$ ratio, draw an arbitrary angle in A with ${\displaystyle {\overline {AB}}}$ as one leg. On the other leg construct ${\displaystyle m+n}$ equidistant points, then draw the line through the last point and B and parallel line through the mth point. This parallel line divides ${\displaystyle {\overline {AB}}}$ in the desired ratio. The graphic to the right shows the partition of a line sgement ${\displaystyle {\overline {AB}}}$ in a ${\displaystyle 5:3}$ ratio.[3]

### Measuring and survey

#### Height of the Cheops pyramid

measuring pieces
computing C and D

According to some historical sources the Greek mathematician Thales applied the intercept theorem to determine the height of the Cheops' pyramid.[1] The following description illustrates the use of the intercept theorem to compute the height of the pyramid. It does not however recount Thales' original work, which was lost.

Thales measured the length of the pyramid's base and the height of his pole. Then at the same time of the day he measured the length of the pyramid's shadow and the length of the pole's shadow. This yielded the following data:

• height of the pole (A): 1.63 m
• shadow of the pole (B): 2 m
• length of the pyramid base: 230 m
• shadow of the pyramid: 65 m

From this he computed

${\displaystyle C=65~{\text{m}}+{\frac {230~{\text{m}}}{2}}=180~{\text{m}}}$

Knowing A,B and C he was now able to apply the intercept theorem to compute

${\displaystyle D={\frac {C\cdot A}{B}}={\frac {1.63~{\text{m}}\cdot 180~{\text{m}}}{2~{\text{m}}}}=146.7~{\text{m}}}$

#### Measuring the width of a river

 The intercept theorem can be used to determine a distance that cannot be measured directly, such as the width of a river or a lake, the height of tall buildings or similar. The graphic to the right illustrates measuring the width of a river. The segments ${\displaystyle |CF|}$,${\displaystyle |CA|}$,${\displaystyle |FE|}$ are measured and used to compute the wanted distance ${\displaystyle |AB|={\frac {|AC||FE|}{|FC|}}}$.

### Parallel lines in triangles and trapezoids

The intercept theorem can be used to prove that a certain construction yields parallel line (segment)s.

 If the midpoints of two triangle sides are connected then the resulting line segment is parallel to the third triangle side. If the midpoints of two the non-parallel sides of a trapezoid are connected, then the resulting line segment is parallel to the other two sides of the trapezoid.

## Proof of the theorem

An elementary proof of the theorem uses triangles of equal area to derive the basic statements about the ratios (claim 1). The other claims then follow by applying the first claim and contradiction.[4]

### Claim 1

 Since ${\displaystyle CA\parallel BD}$, the altitudes of ${\displaystyle \triangle CDA}$ and ${\displaystyle \triangle CBA}$ are of equal length. As those triangles share the same baseline, their areas are identical. So we have ${\displaystyle |\triangle CDA|=|\triangle CBA|}$ and therefore ${\displaystyle |\triangle SCB|=|\triangle SDA|}$ as well. This yields ${\displaystyle {\frac {|\triangle SCA|}{|\triangle CDA|}}={\frac {|\triangle SCA|}{|\triangle CBA|}}}$ and ${\displaystyle {\frac {|\triangle SCA|}{|\triangle SDA|}}={\frac {|\triangle SCA|}{|\triangle SCB|}}}$ Plugging in the formula for triangle areas (${\displaystyle {\tfrac {{\text{baseline}}\cdot {\text{altitude}}}{2}}}$) transforms that into ${\displaystyle {\frac {|SC||AF|}{|CD||AF|}}={\frac {|SA||EC|}{|AB||EC|}}}$ and ${\displaystyle {\frac {|SC||AF|}{|SD||AF|}}={\frac {|SA||EC|}{|SB||EC|}}}$ Canceling the common factors results in: (a) ${\displaystyle \,{\frac {|SC|}{|CD|}}={\frac {|SA|}{|AB|}}}$ and (b) ${\displaystyle \,{\frac {|SC|}{|SD|}}={\frac {|SA|}{|SB|}}}$ Now use (b) to replace ${\displaystyle |SA|}$ and ${\displaystyle |SC|}$ in (a): ${\displaystyle {\frac {\frac {|SA||SD|}{|SB|}}{|CD|}}={\frac {\frac {|SB||SC|}{|SD|}}{|AB|}}}$ Using (b) again this simplifies to: (c) ${\displaystyle \,{\frac {|SD|}{|CD|}}={\frac {|SB|}{|AB|}}}$ ${\displaystyle \,\square }$

### Claim 2

 Draw an additional parallel to ${\displaystyle SD}$ through A. This parallel intersects ${\displaystyle BD}$ in G. Then one has ${\displaystyle |AC|=|DG|}$ and due to claim 1 ${\displaystyle {\frac {|SA|}{|SB|}}={\frac {|DG|}{|BD|}}}$ and therefore ${\displaystyle {\frac {|SA|}{|SB|}}={\frac {|AC|}{|BD|}}}$ ${\displaystyle \square }$

### Claim 3

 Assume ${\displaystyle AC}$ and ${\displaystyle BD}$ are not parallel. Then the parallel line to ${\displaystyle AC}$ through ${\displaystyle D}$ intersects ${\displaystyle SA}$ in ${\displaystyle B_{0}\neq B}$. Since ${\displaystyle |SB|:|SA|=|SD|:|SC|}$ is true, we have ${\displaystyle |SB|={\frac {|SD||SA|}{|SC|}}}$ and on the other hand from claim 2 we have ${\displaystyle |SB_{0}|={\frac {|SD||SA|}{|SC|}}}$. So ${\displaystyle B}$ and ${\displaystyle B_{0}}$ are on the same side of ${\displaystyle S}$ and have the same distance to ${\displaystyle S}$, which means ${\displaystyle B=B_{0}}$. This is a contradiction, so the assumption could not have been true, which means ${\displaystyle AC}$ and ${\displaystyle BD}$ are indeed parallel ${\displaystyle \square }$

### Claim 4

Claim 4 can be shown by applying the intercept theorem for two lines.