مبرهنة طاليس

مبرهنة طاليس في الهندسة تقول أنه إذا قطع مستقيمان 3 مستقيمات متوازية فإننا نحصل على:


\frac{A'B'}{AB}=\frac{B'C'}{BC}=\frac{A'C'}{AC}
Thales theorem 3.png



الصياغة

S is the point the intersection 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 are C, D the intersections of the second line with the 2 parallels such that D is further away from S than C.

  1. The ratios of the any 2 segments on the first line equals the rations 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 2 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 2 intersecting lines are intercepted by 2 arbitrary lines and | SA | : | AB |   =| SC | : | CD |  holds then the 2 intercepting lines are parallel. However the converse of the second statement is not true.
  4. 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.
Intercept theorem 1.jpg


Intercept theorem 2.jpg

التطبيقات

الصياغة الجبرية لمنشآت الفرجار والمسطرة

There are 3 famous problems in elementary geometry, which were posed by the Greek in terms of Compass and straightedge constructions.

  1. Trisecting the angle
  2. Doubling the cube
  3. Squaring the circle

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 2. 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 Intercept theprem multiplication.jpg

Construction of an Inverse Intercept theorem Inverse.jpg

تقسيم خط مستقيم بنسبة معينة

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.

Dividing segment.jpg

القياس/المساحة

Height of the Cheops Pyramid

Thales theorem 6.png

According to some historical sources the Greek mathematician Thales applied the intercept theorem to determine the height of the cheops pyramid. 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

Thales theorem 7.png

قياس عرض النهر

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

River chart.jpg

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.

Triangle midpoints.jpg

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.

Trapezoid midpoints.jpg

Proof

claim 1

Intercept theorem proof.jpg

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  |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

Intercept theorem proof2.jpg

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

Intercept theorem proof3.jpg

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 2 lines.

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