# Teorema dell'intercetta 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.

Contenuti 1 Formulazione 2 Related concepts 2.1 Similarity and similar triangles 2.2 Scalar multiplication in vector spaces 3 Applicazioni 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 Appunti 6 Riferimenti 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: {stile di visualizzazione |Su|:|AB|=|SC|:|CD|} , {stile di visualizzazione |SB|:|AB|=|SD|:|CD|} , {stile di visualizzazione |Su|:|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: {stile di visualizzazione |Su|:|SB|=|SC|:|SD|=|corrente alternata|:|BD|} The converse of the first statement is true as well, cioè. if the two intersecting lines are intercepted by two arbitrary lines and {stile di visualizzazione |Su|:|AB|=|SC|:|CD|} holds then the two intercepting lines are parallel. Tuttavia, 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: {stile di visualizzazione |DI|:|ESSERE|=|FC|:|ED|} , {stile di visualizzazione |DI|:|FC|=|ESSERE|:|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, cioè. 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 particolare {displaystyle lambda cdot ({vec {un}}+{vec {b}})=lambda cdot {vec {un}}+lambda cdot {vec {b}}} e {stile di visualizzazione |lambda {vec {un}}|=|lambda |cdot |{vec {un}}|} ) ensure that the intercept theorem holds. Uno ha {stile di visualizzazione {frac {|lambda cdot {vec {un}}|}{|{vec {un}}|}}={frac {|lambda cdot {vec {b}}|}{|{vec {b}}|}}={frac {|lambda cdot ({vec {un}}+{vec {b}})|}{|{vec {un}}+{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 {stile di visualizzazione a} , a new line segment of length {stile di visualizzazione 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 {stile di visualizzazione {sopra {AB}}} in un {stile di visualizzazione m:n} ratio, draw an arbitrary angle in A with {stile di visualizzazione {sopra {AB}}} as one leg. On the other leg construct {stile di visualizzazione 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 {stile di visualizzazione {sopra {AB}}} in the desired ratio. The graphic to the right shows the partition of a line segment {stile di visualizzazione {sopra {AB}}} in un {stile di visualizzazione 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, però, 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 (UN): 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~{testo{m}}+{frac {230~{testo{m}}}{2}}=180~{testo{m}}} Knowing A,B and C he was now able to apply the intercept theorem to compute {stile di visualizzazione D={frac {Ccdot A}{B}}={frac {1.63~{testo{m}}cdot 180~{testo{m}}}{2~{testo{m}}}}=146.7~{testo{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 {stile di visualizzazione |CF|} , {stile di visualizzazione |circa|} , {stile di visualizzazione |FE|} are measured and used to compute the wanted distance {stile di visualizzazione |AB|={frac {|corrente alternata||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 Notazione: For a triangle the vertical bars ( {stile di visualizzazione |ldot |} ) denote its area and for a line segment its length.

Prova: Da {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 {stile di visualizzazione |triangle CDA|=|triangle CBA|} e quindi {stile di visualizzazione |triangle SCB|=|triangle SDA|} anche. Questo produce {stile di visualizzazione {frac {|triangle SCA|}{|triangle CDA|}}={frac {|triangle SCA|}{|triangle CBA|}}} e {stile di visualizzazione {frac {|triangle SCA|}{|triangle SDA|}}={frac {|triangle SCA|}{|triangle SCB|}}} Plugging in the formula for triangle areas ( {stile di visualizzazione {tfrac {{testo{baseline}}cdot {testo{altitude}}}{2}}} ) transforms that into {stile di visualizzazione {frac {|SC||DI|}{|CD||DI|}}={frac {|Su||EC|}{|AB||EC|}}} e {stile di visualizzazione {frac {|SC||DI|}{|SD||DI|}}={frac {|Su||EC|}{|SB||EC|}}} Canceling the common factors results in: (un) {stile di visualizzazione ,{frac {|SC|}{|CD|}}={frac {|Su|}{|AB|}}} e (b) {stile di visualizzazione ,{frac {|SC|}{|SD|}}={frac {|Su|}{|SB|}}} Now use (b) to replace {stile di visualizzazione |Su|} e {stile di visualizzazione |SC|} in (un): {stile di visualizzazione {frac {frac {|Su||SD|}{|SB|}}{|CD|}}={frac {frac {|SB||SC|}{|SD|}}{|AB|}}} Usando (b) again this simplifies to: (c) {stile di visualizzazione ,{frac {|SD|}{|CD|}}={frac {|SB|}{|AB|}}} {stile di visualizzazione ,square } Claim 2 Draw an additional parallel to {displaystyle SD} through A. This parallel intersects {stile di visualizzazione BD} in G. Then one has {stile di visualizzazione |corrente alternata|=|DG|} and due to claim 1 {stile di visualizzazione {frac {|Su|}{|SB|}}={frac {|DG|}{|BD|}}} e quindi {stile di visualizzazione {frac {|Su|}{|SB|}}={frac {|corrente alternata|}{|BD|}}} {displaystyle quadrato } Claim 3 Assumere {stile di visualizzazione AC} e {stile di visualizzazione BD} are not parallel. Then the parallel line to {stile di visualizzazione AC} attraverso {stile di visualizzazione D} intersects {displaystyle SA} in {stile di visualizzazione B_{0}neq B} . Da {stile di visualizzazione |SB|:|Su|=|SD|:|SC|} è vero, noi abbiamo {stile di visualizzazione |SB|={frac {|SD||Su|}{|SC|}}} and on the other hand from claim 1 noi abbiamo {stile di visualizzazione |SB_{0}|={frac {|SD||Su|}{|SC|}}} .

Così {stile di visualizzazione B} e {stile di visualizzazione B_{0}} are on the same side of {stile di visualizzazione S} and have the same distance to {stile di visualizzazione S} , che significa {displaystyle B=B_{0}} . Questa è una contraddizione, so the assumption could not have been true, che significa {stile di visualizzazione AC} e {stile di visualizzazione BD} are indeed parallel {displaystyle quadrato } Claim 4 Claim 4 can be shown by applying the intercept theorem for two lines.