1. WARM UP EXERCSE A C B b a c h c - x x
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.

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. Early Beginnings
Indeed, the Assyrians had knowledge of the general form before 2000 b.c.
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) 4

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

7. §3-4 Pythagorean Theorem 3
Bhaskara’s dissection proof. c a a b b c c b a a b c
c2 = 4 · ½ · a · b + (b – a)2

8. §3-4 Pythagorean Theorem 4
Garfield’s dissection proof. a b c c a b
½ (a + b) · (a + b) = 2 · ½ · a · b + ½ · c2

10. Extensions
Semicircles
Prove it for homework.

11. Extensions
Golden Rectangles
Prove it for homework.

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 A C B b a c h c - x x
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.

14. Pythagoras Revisited From the previous slide: And of course then,B b a c h
c - x x
From the previous slide:
And of course then,
a 2 + b 2 = cx + c(c – x) = cx + c 2 – cx = c2 14

15. Euclid’s Proof 15

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

17. Ceva’s Theorem
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.
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.

18. Ceva’s Theorem
To prove this theorem we will need the two following theorems.
Theorem – Triangles with the same altitudes have areas in proportion to their bases.
Proven on the next slide.
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: A B N CD
We will prove the if part first.

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

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

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

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

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
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.
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!

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 B N C L M
Note we will use the convention that NB = - BN

30. Menelaus’ Theorem
We will prove the if part first.
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 B N C L M D
AMN  BDN and
BDL  CML by AA

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

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

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

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