# Teorema de interceptação The intercept theorem, also known as Thales's theorem, basic proportionality theorem or side splitter theorem 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. It is traditionally attributed to Greek mathematician Thales. It was known to the ancient Babylonians and Egyptians, although its first known proof appears in Euclid's Elements.

Conteúdo 1 Formulação 2 Related concepts 2.1 Similarity and similar triangles 2.2 Scalar multiplication in vector spaces 3 Formulários 3.1 Algebraic formulation of compass and ruler constructions 3.2 Dividing a line segment in a given ratio 3.3 Measuring and survey 3.3.1 Height of the Cheops pyramid 3.3.2 Measuring the width of a river 3.4 Parallel lines in triangles and trapezoids 4 Prova 4.1 Claim 1 4.2 Claim 2 4.3 Claim 3 4.4 Claim 4 5 Notas 6 Referências 7 External links 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.

The ratio of any two segments on the first line equals the ratio of the according segments on the second line: {estilo de exibição |sobre|:|AB|=|SC|:|CD|} , {estilo de exibição |SB|:|AB|=|SD|:|CD|} , {estilo de exibição |sobre|:|SB|=|SC|:|SD|} The ratio of the two segments on the same line starting at S equals the ratio of the segments on the parallels: {estilo de exibição |sobre|:|SB|=|SC|:|SD|=|CA|:|BD|} The converse of the first statement is true as well, ou seja. if the two intersecting lines are intercepted by two arbitrary lines and {estilo de exibição |sobre|:|AB|=|SC|:|CD|} holds then the two intercepting lines are parallel. No entanto, the converse of the second statement is not true. 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: {estilo de exibição |DO|:|SER|=|FC|:|ED|} , {estilo de exibição |DO|:|FC|=|SER|:|ED|} An example for the case of three lines is given in 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, ou seja. 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 (em particular {displaystyle lambda cdot ({vec {uma}}+{vec {b}})=lambda cdot {vec {uma}}+lambda cdot {vec {b}}} e {estilo de exibição |lambda {vec {uma}}|=|lambda |cdot |{vec {uma}}|} ) ensure that the intercept theorem holds. Um tem {estilo de exibição {fratura {|lambda cdot {vec {uma}}|}{|{vec {uma}}|}}={fratura {|lambda cdot {vec {b}}|}{|{vec {b}}|}}={fratura {|lambda cdot ({vec {uma}}+{vec {b}})|}{|{vec {uma}}+{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: Trisecting the angle Doubling the cube Squaring the circle It took more than 2000 years until all three of them were finally shown to be impossible with the given tools 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 (see constructible number). 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 {estilo de exibição a} , a new line segment of length {estilo de exibição a^{-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 {estilo de exibição {overline {AB}}} em um {estilo de exibição m:n} ratio, draw an arbitrary angle in A with {estilo de exibição {overline {AB}}} as one leg. On the other leg construct {estilo de exibição 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 {estilo de exibição {overline {AB}}} in the desired ratio. The graphic to the right shows the partition of a line segment {estilo de exibição {overline {AB}}} em um {estilo de exibição 5:3} ratio. 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. The following description illustrates the use of the intercept theorem to compute the height of the pyramid. It does not, Contudo, 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 (UMA): 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~{texto{m}}+{fratura {230~{texto{m}}}{2}}=180~{texto{m}}} Knowing A,B and C he was now able to apply the intercept theorem to compute {displaystyle D={fratura {Ccdot A}{B}}={fratura {1.63~{texto{m}}cdot 180~{texto{m}}}{2~{texto{m}}}}=146.7~{texto{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 {estilo de exibição |FC|} , {estilo de exibição |CA|} , {estilo de exibição |FE|} are measured and used to compute the wanted distance {estilo de exibição |AB|={fratura {|CA||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 (Midpoint theorem of triangles).

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

Proof 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. Claim 1 Notação: For a triangle the vertical bars ( {estilo de exibição |ldots |} ) denote its area and for a line segment its length.

Prova: Desde {displaystyle CAparallel BD} , the altitudes of {displaystyle triangle CDA} e {displaystyle triangle CBA} are of equal length. As those triangles share the same baseline, their areas are identical. So we have {estilo de exibição |triangle CDA|=|triangle CBA|} e, portanto, {estilo de exibição |triangle SCB|=|triangle SDA|} também. Isso rende {estilo de exibição {fratura {|triangle SCA|}{|triangle CDA|}}={fratura {|triangle SCA|}{|triangle CBA|}}} e {estilo de exibição {fratura {|triangle SCA|}{|triangle SDA|}}={fratura {|triangle SCA|}{|triangle SCB|}}} Plugging in the formula for triangle areas ( {estilo de exibição {tfrac {{texto{baseline}}cdot {texto{altitude}}}{2}}} ) transforms that into {estilo de exibição {fratura {|SC||DO|}{|CD||DO|}}={fratura {|sobre||EC|}{|AB||EC|}}} e {estilo de exibição {fratura {|SC||DO|}{|SD||DO|}}={fratura {|sobre||EC|}{|SB||EC|}}} Canceling the common factors results in: (uma) {estilo de exibição ,{fratura {|SC|}{|CD|}}={fratura {|sobre|}{|AB|}}} e (b) {estilo de exibição ,{fratura {|SC|}{|SD|}}={fratura {|sobre|}{|SB|}}} Now use (b) to replace {estilo de exibição |sobre|} e {estilo de exibição |SC|} dentro (uma): {estilo de exibição {fratura {fratura {|sobre||SD|}{|SB|}}{|CD|}}={fratura {fratura {|SB||SC|}{|SD|}}{|AB|}}} Using (b) again this simplifies to: (c) {estilo de exibição ,{fratura {|SD|}{|CD|}}={fratura {|SB|}{|AB|}}} {estilo de exibição ,square } Claim 2 Draw an additional parallel to {displaystyle SD} through A. This parallel intersects {estilo de exibição BD} in G. Then one has {estilo de exibição |CA|=|DG|} and due to claim 1 {estilo de exibição {fratura {|sobre|}{|SB|}}={fratura {|DG|}{|BD|}}} e, portanto, {estilo de exibição {fratura {|sobre|}{|SB|}}={fratura {|CA|}{|BD|}}} {quadrado de estilo de exibição } Claim 3 Presumir {estilo de exibição AC} e {estilo de exibição BD} are not parallel. Then the parallel line to {estilo de exibição AC} Através dos {estilo de exibição D} intersects {displaystyle SA} dentro {estilo de exibição B_{0}neq B} . Desde {estilo de exibição |SB|:|sobre|=|SD|:|SC|} é verdade, temos {estilo de exibição |SB|={fratura {|SD||sobre|}{|SC|}}} and on the other hand from claim 1 temos {estilo de exibição |SB_{0}|={fratura {|SD||sobre|}{|SC|}}} .

Então {estilo de exibição B} e {estilo de exibição B_{0}} are on the same side of {estilo de exibição S} and have the same distance to {estilo de exibição S} , que significa {displaystyle B=B_{0}} . Isso é uma contradição, so the assumption could not have been true, que significa {estilo de exibição AC} e {estilo de exibição BD} are indeed parallel {quadrado de estilo de exibição } Claim 4 Claim 4 can be shown by applying the intercept theorem for two lines.