1. WARM UP EXERCSE the altitude to the right angle of a right triangle forms two new right triangles which are similar to the original right triangle. List all the proportions you can among these three triangles. A C B b a c h c - xx

2. Early Beginnings In ancient times the special relationship between a right triangle and the squares on the three sides was known. 2

3. Early Beginnings OR 3

4. Indeed, the Assyrians had knowledge of the general form before 2000 b.c. 4 Early Beginnings The Babylonians had knowledge of all of the Pythagorean triples and had a formula to generate them. ( 3, 4, 5 )( 5, 12, 13)( 7, 24, 25)( 8, 15, 17) ( 9, 40, 41)(11, 60, 61)(12, 35, 37)(13, 84, 85) (16, 63, 65)(20, 21, 29)(28, 45, 53)(33, 56, 65) (36, 77, 85)(39, 80, 89)(48, 55, 73)(65, 72, 97)

5. 5

6. 6 §3-4 Pythagorean Theorem 2 Pythagorean dissection proof. a a a a b b b b c c c c c = a a a a b b b b

7. 7 §3-4 Pythagorean Theorem 3 Bhaskara’s dissection proof. a a a b b b b a c c c c c 2 = 4 · ½ · a · b + (b – a) 2

8. 8 §3-4 Pythagorean Theorem 4 Garfield’s dissection proof. a b b a c c ½ (a + b) · (a + b) = 2 · ½ · a · b + ½ · c 2

9. 9 http://www.usna.edu/MathDept/mdm/pyth.html

10. 10 Extensions Semicircles Prove it for homework.

11. 11 Extensions Golden Rectangles Prove it for homework.

12. 12 THE GENERAL EXTENSION TO PYTHAGORAS' THEOREM: If any 3 similar shapes are drawn on the sides of a right triangle, then the area of the shape on the hypotenuse equals the sum of the areas on the other two sides.

13. WARMU UP EXERCSE the altitude to the right angle of a right triangle forms two new right triangles which are similar to the original right triangle. List all the proportions you can among these three triangles. A C B b a c h c - xx

14. Pythagoras Revisited 14 From the previous slide: A C B b a c h c - x x And of course then, a 2 + b 2 = cx + c(c – x) = cx + c 2 – cx = c 2

15. 15 Euclid’s Proof http://www.cut-the- knot.org/pythagoras/more y.shtml

16. 16 First Assignment Find another proof of the Pythagorean Theorem.

17. Ceva’s Theorem Definition – A Cevian is a line from a vertex of a triangle through the opposite side. Altitudes, medians and angle bisectors are examples of Cevians. The theorem is often attributed to Giovanni Ceva, who published it in his 1678 work De lineis rectis. But it was proven much earlier by Yusuf Al-Mu'taman ibn Hűd, an eleventh-century king of Zaragoza.

18. Ceva’s Theorem Theorem – Triangles with the same altitudes have areas in proportion to their bases. Proven on the next slide. To prove this theorem we will need the two following theorems. Theorem Proof as homework.

19. Ceva’s Theorem We will be using the fact that triangles that have the same altitudes have areas in proportion to their bases. A N C B h

20. Ceva’s Theorem Three Cevians concur iff the following is true: We will prove the if part first. ABN C L MD

21. Ceva’s Theorem We will be using the fact that triangles that have the same altitudes have areas in proportion to their bases and the following three ratios to prove our theorem. We will begin with the ration AN/NB.

22. Ceva’s Theorem 22 Given: AL, BM, CN concur Theorem. Theorem (1) & (2) Transitive What is given? What will we prove? Why? Ratio property and subtraction of areas. Look at what we have just shown. ABN C L M D Prove: (1) (2) (3) (4) ABN C L M D D Why Continued

23. Ceva’s Theorem We just saw that: Using the same argument: ABN C L M D D (4) ABN C L M D (5)(6) Thus

24. Ceva’s Theorem We now move to the only if part.

25. Ceva’s Theorem 25 First half Ceva’s Theorem Given. (1) & (2) Transitive What is given? What will we prove? Why? Simplification. Which is true only if N’ = N Prove: AN, BM, CL concur Given: (1) (2) (3) (4) Why Assume AL and BM intersect at D and the other line through D is CN’. AB N C L M D H

26. Ceva’s Theorem This is an important theorem and can be used to prove many theorems where you are to show concurrency.

27. Ceva’s Theorem There is also a proof of this theorem using similar triangles. If you want it I will send you a copy.

28. Menelaus’ Theorem This is the duel of Ceva’s Theorem. Whereas Ceva’s Theorem is used for concurrency, Menelaus’ Theorem is used for colinearity of three points. AND its proof is easier! Menelaus' theorem, named for Menelaus (70-130ad) of Alexandria. Very little is known about Menelaus's life, it is supposed that he lived in Rome, where he probably moved after having spent his youth in Alexandria. He was called Menelaus of Alexandria by both Pappus of Alexandria and Proclus.

29. Menelaus’ Theorem Given points A, B, C that form triangle ABC, and points L, M, N that lie on lines BC, AC, AB, then L, M, N are collinear if and only if A BN C L M Note we will use the convention that NB = - BN

30. Menelaus’ Theorem Begin by constructing a line parallel to AC through B. It will intersect MN in a new point D producing two sets of similar triangles. A BN C L M D We will prove the if part first.  AMN   BDN and  BDL   CML by AA

31. Menelaus’ Theorem 31 Given: M, L, N, Collinear Previous slide. (1) · (2) What is given? What will we prove? Why? Simplify Prove: (1) (2) (3) (4) Why A BN C L M D (5) Why MA = - AM QED

32. Menelaus’ Theorem We now move to the only if part. Given Show that L, M, and N are collinear.

33. Menelaus’ Theorem 33 First half Menelaus’ Theorem Given. (1) & (2) Transitive What is given? What will we prove? Why? Simplification. Which is true only if N’ = N Prove: L, M, N collinear. Given: (1) (2) (3) (4) Why Assume L, M, N’ ≠ N collinear. A BN’ C L M

34. Wrap-up We looked at several proofs and some history of the Pythagorean Theorem. We proved Ceva’s Theorem. We proved Menelaus’ Theorem.

35. Next Class We will cover the lesson Transformations 1.

36. Second Assignment Learn the proofs for Ceva’s and Menelaus’ theorems.