Intercept theorem

# Intercept theorem

Discussion

Encyclopedia
{{distinguish|Thales' theorem}} The intercept theorem, also known as Thales' theorem (not to be confused with another theorem with that name
Thales' theorem
In geometry, Thales' theorem states that if A, B and C are points on a circle where the line AC is a diameter of the circle, then the angle ABC is a right angle. Thales' theorem is a special case of the inscribed angle theorem...

), is an important theorem in elementary geometry about the ratios of various line segment
Line segment
In geometry, a line segment is a part of a line that is bounded by two end points, and contains every point on the line between its end points. Examples of line segments include the sides of a triangle or square. More generally, when the end points are both vertices of a polygon, the line segment...

s that are created if two intersecting line
Line (geometry)
The notion of line or straight line was introduced by the ancient mathematicians to represent straight objects with negligible width and depth. Lines are an idealization of such objects...

s are intercepted by a pair of parallel
Parallel (geometry)
Parallelism is a term in geometry and in everyday life that refers to a property in Euclidean space of two or more lines or planes, or a combination of these. The assumed existence and properties of parallel lines are the basis of Euclid's parallel postulate. Two lines in a plane that do not...

s. It is equivalent to the theorem about ratios in similar triangles. Traditionally it is attributed to Greek mathematician Thales
Thales
Thales of Miletus was a pre-Socratic Greek philosopher from Miletus in Asia Minor, and one of the Seven Sages of Greece. Many, most notably Aristotle, regard him as the first philosopher in the Greek tradition...

.

## 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. NEWLINE
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1. The ratios of any two segments on the first line equals the ratios of the according segments on the second line: $| SA | : | AB | =| SC | : | CD |$, $| SB | : | AB | =| SD | : | CD |$, $| SA | : | SB | =| SC | : | SD |$
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3. The ratio of the two segments on the same line starting at S equals the ratio of the segments on the parallels: $| SA |:| SB | = | SC | :| SD | =| AC | : | BD |$
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5. 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.
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7. 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.
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### Similarity and similar Triangles

The intercept theorem is closely related to similarity
Similarity (geometry)
Two geometrical objects are called similar if they both have the same shape. More precisely, either one is congruent to the result of a uniform scaling of the other...

. 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
Converse (logic)
In logic, the converse of a categorical or implicational statement is the result of reversing its two parts. For the implication P → Q, the converse is Q → P. For the categorical proposition All S is P, the converse is All P is S. In neither case does the converse necessarily follow from...

the intercept theorem configuration always contains two similar triangles.

### Scalar Multiplication in Vector Spaces

In a normed vector space
Vector space
A vector space is a mathematical structure formed by a collection of vectors: objects that may be added together and multiplied by numbers, called scalars in this context. Scalars are often taken to be real numbers, but one may also consider vector spaces with scalar multiplication by complex...

, the axiom
Axiom
In traditional logic, an axiom or postulate is a proposition that is not proven or demonstrated but considered either to be self-evident or to define and delimit the realm of analysis. In other words, an axiom is a logical statement that is assumed to be true...

s concerning the scalar multiplication
Scalar multiplication
In mathematics, scalar multiplication is one of the basic operations defining a vector space in linear algebra . In an intuitive geometrical context, scalar multiplication of a real Euclidean vector by a positive real number multiplies the magnitude of the vector without changing its direction...

(in particular $\lambda \cdot \left(\vec\left\{a\right\}+\vec\left\{b\right\}\right)=\lambda \cdot \vec\left\{a\right\}+ \lambda \cdot \vec\left\{b\right\}$ and $\|\lambda \vec\left\{a\right\}\|=|\lambda|\cdot\ \|\vec\left\{a\right\}\|$) are assuring that the intercept theorem holds. You have $\frac\left\{ \| \lambda \cdot \vec\left\{a\right\} \| \right\}\left\{ \| \vec\left\{a\right\} \|\right\}$

### 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.NEWLINE
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1. Trisecting the angle
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3. Doubling the cube
Doubling the cube
Doubling the cube is one of the three most famous geometric problems unsolvable by compass and straightedge construction...

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5. Squaring the circle
Squaring the circle
Squaring the circle is a problem proposed by ancient geometers. It is the challenge of constructing a square with the same area as a given circle by using only a finite number of steps with compass and straightedge...

NEWLINE 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 extension
Field extension
In abstract algebra, field extensions are the main object of study in field theory. The general idea is to start with a base field and construct in some manner a larger field which contains the base field and satisfies additional properties...

s, one needs to match field operations
Field (mathematics)
In abstract algebra, a field is a commutative ring whose nonzero elements form a group under multiplication. As such it is an algebraic structure with notions of addition, subtraction, multiplication, and division, satisfying certain axioms...

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 $d^\left\{-1\right\}$. The intercept theorem can be used to show that in both cases the construction is possible. NEWLINENEWLINENEWLINENEWLINENEWLINENEWLINE
 Construction of a product Construction of an Inverse
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### Dividing a line segment in a given ratio

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 To divide an arbitrary line segment $\overline\left\{AB\right\}$ in a $m:n$ ratio, draw an arbitrary angle in A with $\overline\left\{AB\right\}$ as one leg. One 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 $\overline\left\{AB\right\}$ in the desired ratio. The graphic to the right shows the partition of a line sgement $\overline\left\{AB\right\}$ in a $5:3$ ratio.
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#### Height of the Cheops Pyramid

According to some historical sources the Greek mathematician Thales
Thales
Thales of Miletus was a pre-Socratic Greek philosopher from Miletus in Asia Minor, and one of the Seven Sages of Greece. Many, most notably Aristotle, regard him as the first philosopher in the Greek tradition...

applied the intercept theorem to determine the height of the Cheops' pyramid
Great Pyramid of Giza
The Great Pyramid of Giza is the oldest and largest of the three pyramids in the Giza Necropolis bordering what is now El Giza, Egypt. It is the oldest of the Seven Wonders of the Ancient World, and the only one to remain largely intact...

. 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 yieldedthe following data:NEWLINE
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• height of the pole (A): 1.63m
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• shadow of the pole (B): 2m
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• length of the pyramid base: 230m
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• shadow of the pyramid: 65m
NEWLINE From this he computed$C = 65m+\frac\left\{230m\right\}\left\{2\right\}=180m$ Knowing A,B and C he was now able to apply the intercept theorem to compute$D=\frac\left\{C \cdot A\right\}\left\{B\right\}=\frac\left\{1.63m \cdot 180m\right\}\left\{2m\right\}=146.7m$

#### Measuring the Width of a River

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 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, 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 $|AB|=\frac\left\{|AC$ }{|FC|} .
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### Parallel Lines in Triangles and Trapezoids

The intercept theorem can be used to prove that a certain construction yields parallel line (segment)s. NEWLINENEWLINENEWLINENEWLINENEWLINENEWLINE
 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.
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### claim 1

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 Due to heights of equal length ($CA\parallel BD$ ) we have $| \triangle CDA|=| \triangle CBA|$ and therefore $| \triangle SCB|=| \triangle SDA|$. This yields $\frac\left\{| \triangle SCA|\right\}\left\{|\triangle CDA|\right\}=\frac\left\{|\triangle SCA|\right\}\left\{|\triangle CBA|\right\}$ and $\frac\left\{| \triangle SCA|\right\}\left\{|\triangle SDA|\right\}=\frac\left\{|\triangle SCA|\right\}\left\{|\triangle SCB|\right\}$ Plugging in the actual formula for triangle areas ($\frac\left\{baseline \cdot height\right\}\left\{2\right\}$) transforms that into $\frac\left\{|SC$ }{|CD }=\frac{|SA }{|AB } and $\frac\left\{|SC$ }{|SD }=\frac{|SA }{|SB } Cancelling the common factors results in: (a) $\, \frac\left\{|SC|\right\}\left\{|CD|\right\}=\frac\left\{|SA|\right\}\left\{|AB|\right\}$ and (b) $\, \frac\left\{|SC|\right\}\left\{|SD|\right\}=\frac\left\{|SA|\right\}\left\{|SB|\right\}$ Now use (b) to replace $|SA|$ and $|SC|$ in (a): $\frac\left\{\frac\left\{|SA$ }{|SB|}}{|CD|}=\frac{\frac{|SB }{|SD|}}{|AB|} Using (b) again this simplifies to: (c) $\, \frac\left\{|SD|\right\}\left\{|CD|\right\}=\frac\left\{|SB|\right\}\left\{|AB|\right\}$ $\, \square$
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### claim 2

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 Draw an additional parallel to $SD$ through A. This parallel intersects $BD$ in G. Then you have $|AC|=|DG|$ and due to claim 1 $\frac\left\{|SA|\right\}\left\{|SB|\right\}=\frac\left\{|DG|\right\}\left\{|BD|\right\}$ and therefore $\frac\left\{|SA|\right\}\left\{|SB|\right\}=\frac\left\{|AC|\right\}\left\{|BD|\right\}$ $\square$
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### claim 3

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