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.

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

Early Beginnings OR 3

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

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

§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

§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

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

Extensions Semicircles Prove it for homework.

Extensions Golden Rectangles Prove it for homework.

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.

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.

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

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

First Assignment Find another proof of the Pythagorean Theorem.

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.

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.

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

Ceva’s Theorem Three Cevians concur iff the following is true: A B N C D We will prove the if part first.

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.

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

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

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

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

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

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

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!

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

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

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

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

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

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

Next Class We will cover the lesson Transformations 1.

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