
#1
Jan3012, 10:58 PM

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CD is a perpendicular from C to AB. Prove that triangles ACD and CBD are both simlar to triangle ABC. (See attached image)
I can prove corresponding angles are congruent very easily for ABC and CBD. For example, angle A = angle A for both ABC and ACD. Also, angle D = angle C since ABC is a right triangle and CD is perpendicular to AB. Lastly, angle ACD = 90  angle DAC and angle ABC = 90  angle DAC. Therefore, angle ACD = angle ABC. Thus all three corresponding angles are congruent. For showing that the sides are proportional, I have no clue. What confuses me is what c represents. I am guessing it is supposed to represent AB. If it does, then what in the world is AD supposed to be? Or if c represents AD then what is DB supposed to be? Thanks for all your help. Tell me if I have started on the wrong foot or if I am way off in my approach. Any feedback is welcome. 



#2
Jan3112, 07:28 AM

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Hi DEMJR! Welcome to PF!




#3
Jan3112, 07:52 AM

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Similar triangles is equivalent to the Pythagorean theorem. When two triangles have the same angles then their corresponding sides have the same ratios.
In Euclidean geometry shape is scale invariant. This is not true of hyperbolic and spherical geometry. The proof you want to try is to show that the Pythagorean Theorem implies similar triangles and the converse, that similar triangle implies the Pythagorean theorem. Also show that the parallel postulate is true only when these are true, 



#4
Jan3112, 12:57 PM

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Prove similarity (leading to a Pyth. proof) 



#5
Jan3112, 01:09 PM

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You have to define what similarity means. I.e. if AB, CD, XY, ZW are line segments, you have to define what the equation AB/CD = XY/ZW. Equivalently, you have to define the equation (AB)(ZW) = (XY)(CD), i.e. you have to define an equality of products of line segments.
One way to do this is to say that a product of line segments represents the area of the rectangle with the segments as sides. Then you have to prove that the rectangles with sides (AB) and (ZW) has the same area as the one with sides (XY) and (CD). When these segments represent corresponding sides of a triangle with the same angles, one way to prove this is to note that the triangles can be placed in a circle so that they are produced by two intersecting secants of the circle. Then Proposition III.35 of Euclid implies the rectangles they determine have the "same area", in the sense of equi  decomposition. This proof uses pythagoras, as lavinia has said. Euclid's own proof is different and makes an additional assumption, that any two segments considered can be used to measure each other. I.e. that there exist integers n,m such that n times one segment exceeds the other, and m times the other exceeds the one. This is called Archimedes axiom, although Euclid specifically mentions its necessity. Then since we know what it means to multiply a segment by an integer, Euclid defines in chapter 5, the equation (AB)/(CD) = (XY)/(ZW) to mean that for every pair of integers n,m we have (AB)/(CD) ≤ n/m if and only if (XY)/(ZW) ≤ n/m. Of course here, (AB)/(CD) ≤ n/m, means that m(AB) ≤ n(CD). This last statement makes sense because we know how to multiply a segment by an integer. Euclid then painstakingly uses the theory of area to prove similarity in this sense in Proposition VI.4. Then he proves the result you ask about in Prop. VI.8. The fact he does not use the earlier result of Prop. III.35 to obtain similarity in an even stronger sense than here makes me wonder if the same person who wrote book III, also wrote books V and VI. It is often asserted that the theory of similarity should be sued to prove pythagoras, but this is circular in some presentations, since in Euclid pythagoras is used to develop the theory of similarity. Indeed similarity can be developed without pythagoras, but it is a bit of trouble to do so. This is achieved nicely in Hartshorne's geometry, euclid and beyond. After using Pythgoras to develop similarity Euclid does give a second proof of pythagoras using similarity. 



#6
Jan3112, 08:22 PM

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Wikipedia says that if the angles of a triangle are equal then that is one of the definitions of similar. This must be your answer because without some theorem such as the Pythagorean theorem you can not say anything about the ratios of the corresponding sides.
http://en.wikipedia.org/wiki/Similar...ilar_triangles 



#7
Feb112, 04:43 AM

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see also proposition iv of book vi of euclid's elements …
"the sides about the equal angles of equiangular triangles are proportionals" (eg page 172 of http://books.google.co.uk/books?id=6...ok+VI%22&hl=en ) 



#8
Feb112, 01:00 PM

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BTW: in this problem you are assuming that the sum of the angles of a triangles is 180 degrees. That is only true if the Pythagorean Theorem is true.




#9
Feb112, 10:00 PM

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That wikipedia article tacitly assumes it is discussing euclidean geometry and not some other geometry. the phenomenon of scaling is peculiar to the euclidean case. even there, it must be proved that the two properties are equivalent, equal angles and proportional sides. Moreover the meaning of the concept of proportional sides must be defined. Remember that there are no real numbers in euclid, so it is meaningless just to say the lengths of the sides are proportional. euclid does have a concept of "equal area" without using numbers, by considering two figures to be equal (in area) if they are equidecomposable, or are the difference of two such figures. Thus I have used this concept above to define proportional sides without using numbers. If one does use numbers, one can prove the basic similarity result more easily, as i outline here.
just assuming the parallel postulate and nothing about areas, one can show e.g. that a line that bisects one side of a triangle, and is parallel to the base, also bisects the other side. This is the first case of the theorem on similarity. I.e. this implies the fact that a triangle with the same angles as another triangle, and with one side equal to half the corresponding side of the first triangle, in fact has every side equal to half the corresponding side of the first triangle. Thus the triangles are similar, in the sense of having proportional sides. Next we prove the same result for triangles with the same angles and such that one side is 1/3 the corresponding side of the first triangle. Then we do it for triangles with the same angles, and one side any rational multiple of the first triangle. Then using the fact that rational numbers are dense in all real numbers we deduce the result for all triangles with the same angles, under the assumption of the archimedean axiom for our geometry. This proof does not use pythagoras, but of course uses the equivalent parallel postulate. I attach it from my class notes from 2007. 


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