Intercept theorem

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The intercept theorem is an important theorem in elementary geometry about the ratios of various line segments, that are created if 2 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, which is the reason why it is named theorem of Thales in some languages.

[edit] Formulation

S is the intersection point of 2 lines and A, B are the intersections of the first line with the 2 parallels, such that B is further away from S than A, and similarly C, D are the intersections of the second line with the 2 parallels such that D is further away from S than C.

The ratios of any 2 segments on the first line equals the ratios of the according segments on the second line: $| S A | : | A B | = | S C | : | C D |$ , $| S B | : | A B | = | S D | : | C D |$ , $| S A | : | S B | = | S C | : | S D |$
The ratio of the 2 segments on the same line starting at S equals the ratio of the segments on the parallels: $| S A | : | S B | = | S C | : | S D | = | A C | : | B D |$
The converse of the first statement is true as well, i.e. if the 2 intersecting lines are intercepted by 2 arbitrary lines and $| S A | : | A B | = | S C | : | C D |$ holds then the 2 intercepting lines are parallel. However the converse of the second statement is not true.
If you have more than 2 lines intersecting in S, then ratio of the 2 segments on a parallel equals the ratio of the according segments on the other parallel. An example for the case of 3 lines is given the second graphic below.

[edit] Related Concepts

[edit] Similarity and similar Triangles

Arranging 2 similar triangles, so that the intercept theorem can be applied

The intercept theorem is closely related to the 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 2 similar triangles in one another, so that you get the configuration in which the intercepts applies and vice versa the intercept theorem configuartion contains always 2 similar triangles.

[edit] 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 $\frac{ \| \lambda \cdot \vec{a} \| }{ \| \vec{a} \|} =\frac{\|\lambda\cdot\vec{b}\|}{\|\vec{b}\|} =\frac{\|\lambda\cdot(\vec{a}+\vec{b}) \|}{\|\vec{a}+\vec{b}\|} =|\lambda|$

[edit] Applications

[edit] Algebraic formulation of Compass and Ruler Constructions

There are 3 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 3 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 2 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 the construction is possible.

Construction of a product	Construction of an Inverse

[edit] Dividing a line segment in a given ratio

To divide an arbitrary line segment $\overline{AB}$ in a $m : n$ ratio you draw an arbitrary angle in A with $\overline{AB}$ as one leg. One other leg you construct $m + n$ equidistant points, then you draw 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.

[edit] Measuring/Survey

[edit] Height of the Cheops Pyramid

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 however not recount Thales' original work, which was lost.

He measured the shadow of a pole (B) with 2m, its height (A) with 1.63m, the shadow of the cheops pyramid with 63m, its base length with 230m. From that he computed

$C = 2m+63m+\frac{230m}{2}=180m$

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

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

[edit] Measuring the Width of a River

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

[edit] Parallel Lines in Triangles and Trapezoids

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

If the midpoints of 2 triangle sides are connected then the resulting line segment is parallel to the 3rd triangle side.	If the midpoints of 2 the non parallel sides of a trapezoid are connected, then the result line segment is parallel to the other 2 sides of the trapezoid.

[edit] Proof

[edit] claim 1

Due to heights of equal length ( $CA\parallel BD$ ) we have $\| \triangle CDA\|=\| \triangle CBA\|$ and therefore $\| \triangle SCB\|=\| \triangle SDA\|$ . 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 actual formula for triangle areas ( $\frac{baseline \cdot height}{2}$ ) transforms that into $\frac{\|SC\|\|AF\|}{\|CD\|\|AF\|}=\frac{\|SA\|\|EC\|}{\|AB\|\|EC\|}$ and $\frac{\|SC\|\|AF\|}{\|SD\|\|AF\|}=\frac{\|SA\|\|EC\|}{\|SB\|\|EC\|}$ Cancelling 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 $\| S A \|$ and $\| S C \|$ in (a): $\frac{\frac{\|SA\|\|SD\|}{\|SB\|}}{\|CD\|}=\frac{\frac{\|SB\|\|SC\|}{\|SD\|}}{\|AB\|}$ Using (b) again this simplifies to: (c) $\, \frac{\|SD\|}{\|CD\|}=\frac{\|SB\|}{\|AB\|}$ $\, \square$

[edit] claim 2

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

[edit] claim 3

Assume $A C$ and $B D$ are not parallel. Then the parallel line to $A C$ through $D$ intersects $S A$ in $B_{0}\neq B$ . Since $\| S B \| : \| S A \| = \| S D \| : \| S C \|$ 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 $A C$ and $B D$ are indeed parallel $\square$

[edit] claim 4

Can be shown by applying the intercept theorem for 2 lines.

[edit] See also

[edit] Notes

^ No original work of Thales has survived. All historical sources that attribute the intercept theorem or related knowledge to him were written centuries after his death. Diogenes Laertius and Pliny give description that strictly speaking does not require the intercept theorem but can rely on a simple observation only, namely that at certain point of the day the length of an object's shadow will match its height. Laertius quotes a statement of the philosopher Hieronymus about Thales: "Hieronymus says that [Thales] even succeeded in measuring the pyramids by observation of the length of their shadow at the moment when our shadows are equal to our own height.". Pliny writes: "Thales discovered how to obtain the height of pyramids and all other similar objects, namely, by measuring the shadow of the object at the time when a body and its shadow are equal in length.". However Plutarch gives an account, that may suggest Thales knowing the intercept theorem or at least a special case of it:".. without trouble or the assistance of any instrument [he] merely set up a stick at the extremity of the shadow cast by the pyramid and, having thus made two triangles by the impact of the sun's rays, ... showed that the pyramid has to the stick the same ratio which the shadow [of the pyramid] has to the shadow [of the stick]". (Source: Thales biography of the MacTutor).

[edit] References

Schupp, H.: Elementargeometrie. UTB Schöningh 1977, ISBN 3-506-99189-2 (German)

[edit] External links

Intercept theorem at mathsrevision.net
Java-Applet illustrating the intercept theorem
Thales biography at the MacTutor

[0] No original work of Thales has survived. All historical sources that attribute the intercept theorem or related knowledge to him were written centuries after his death. Diogenes Laertius and Pliny give description that strictly speaking does not require the intercept theorem but can rely on a simple observation only, namely that at certain point of the day the length of an object's shadow will match its height. Laertius quotes a statement of the philosopher Hieronymus about Thales: "Hieronymus says that [Thales] even succeeded in measuring the pyramids by observation of the length of their shadow at the moment when our shadows are equal to our own height.". Pliny writes: "Thales discovered how to obtain the height of pyramids and all other similar objects, namely, by measuring the shadow of the object at the time when a body and its shadow are equal in length.". However Plutarch gives an account, that may suggest Thales knowing the intercept theorem or at least a special case of it:".. without trouble or the assistance of any instrument [he] merely set up a stick at the extremity of the shadow cast by the pyramid and, having thus made two triangles by the impact of the sun's rays, ... showed that the pyramid has to the stick the same ratio which the shadow [of the pyramid] has to the shadow [of the stick]". (Source: Thales biography of the MacTutor).

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