By Otto Schreier, Emanuel Sperner
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Extra info for Projective Geometry of N Dimensions (Volume Two of Introduction to Modern Algebra and Matrix Theory)
26 Thales’ Theorem, Case 2 Several beautiful problems on the application of Thales’ Theorem will be presented in the “Construction” section. Problem 12. Prove the Pythagorean Theorem using relations between similar triangles formed by the height dropped from the vertex of the right angle of the right triangle. Proof. Actually this height dropped from a 90 angle is a very important auxiliary element! For example, as it is shown in Fig. 27, triangles DCB and ABC are similar, then the following is true C a b h B x O D A y Fig.
Since BO1 and BO2 are bisectors of angles ABD and DBC, respectively, and using the condition of the problem we get, mﬀ O2 BD ¼ 2α. Consider triangles BO2N and BO1M. Using the double angle formula for tan 2α we obtain: BN ¼ 8 8ð1 À tan2 αÞ 4 ¼ ¼ BM < tan 2α 2 tan2 α tan α and jDMj < jDNj. Since |DF| ¼ |DM| and |DN| ¼ |DE| (property of a tangent to a pﬃﬃﬃﬃﬃﬃﬃﬃ circle), then jDMj þ jDNj ¼ jDEj þ jDFj ¼ jEFj ¼ 129. B N O2 O1 M A j F D C E Fig. 30 Sketch for Problem 15 Since O2D is the bisector of angle CDB, O1D is the bisector of angle ADB, and angles ADM and NDE are supplementary, we have that mﬀO1 DO2 ¼ 90 .
The ratio between these sides can be written as BD/CD. 4 Similar Triangles 23 From which we can conclude that CD is the geometric mean (the square root of the product of the components) of the sides DA and BD. See more on this topic in Sect. 1. 1 Thales’ Theorem Similar triangles and their properties were known to Egyptians and Babylonians. However, we need to give credit to the famous Greek geometer Thales who first formulated the properties of similar triangles formed by two parallel lines and two intersecting transversals.