# 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.

## 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: $| SA | : | AB | =| SC | : | CD |$, $| SB | : | AB | =| SD | : | CD |$, $| SA | : | SB | =| SC | : | SD |$
2. The ratio of the two segments on the same line starting at S equals the ratio of the segments on the parallels: $| SA |:| SB | = | SC | :| SD | =| AC | : | BD |$
3. The converse of the first statement is true as well, i.e. if the two intersecting lines are intercepted by two arbitrary lines and $| SA | : | AB | =| SC | : | CD |$ holds then the two intercepting lines are parallel. However the converse of the second statement is not true.
4. 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. An example for the case of three lines is given the second graphic below.

## 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. In fact 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 $\lambda \cdot (\vec{a}+\vec{b})=\lambda \cdot \vec{a}+ \lambda \cdot \vec{b}$ and $\|\lambda \vec{a}\|=|\lambda|\cdot\ \|\vec{a}\|$) are assuring that the intercept theorem holds. You have $\frac{ \| \lambda \cdot \vec{a} \| }{ \| \vec{a} \|} =\frac{\|\lambda\cdot\vec{b}\|}{\|\vec{b}\|} =\frac{\|\lambda\cdot(\vec{a}+\vec{b}) \|}{\|\vec{a}+\vec{b}\|} =|\lambda|$

## Applications

### Algebraic formulation of Compass and Ruler Constructions

There are three famous problems in elementary geometry which were posed by the Greek in terms of Compass and straightedge constructions.

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 $d$, a new line segment of length $d^{-1}$. 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

### Dividing a line segment in a given ratio

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

### Measuring/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 Cheops' 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): 1.63m
• shadow of the pole (B): 2m
• length of the pyramid base: 230m
• shadow of the pyramid: 65m

From this he computed

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

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

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

#### 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 $|CF|$,$|CA|$,$|FE|$ are measured and used to compute the wanted distance $|AB|=\frac{|AC||FE|}{|FC|}$.

### 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

### claim 1

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

### claim 2

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

### claim 3

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

### claim 4

Can be shown by applying the intercept theorem for two lines.