Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Steinitz's theorem wikipedia , lookup
Euler angles wikipedia , lookup
Noether's theorem wikipedia , lookup
Riemann–Roch theorem wikipedia , lookup
Reuleaux triangle wikipedia , lookup
Rational trigonometry wikipedia , lookup
Trigonometric functions wikipedia , lookup
History of trigonometry wikipedia , lookup
Brouwer fixed-point theorem wikipedia , lookup
Four color theorem wikipedia , lookup
Euclidean geometry wikipedia , lookup
Theorem 1: The sum of the degree measures of the angles of a triangle is 1800 Given: Triangle To Prove: ∠1 + ∠2 + ∠3 = 1800 Construction: Draw line through ∠3 parallel to the base 4 3 5 Proof: ⇒ 1 2 ∠3 + ∠4 + ∠5 = 1800 Straight line ∠1 = ∠4 and ∠2 = ∠5 Alternate angles ∠3 + ∠1 + ∠2 = 1800 ∠1 + ∠2 + ∠3 = 1800 Q.E.D. Theorem 2: The opposite sides of a parallelogram have equal lengths. Given: Parallelogram abcd To Prove: |ab| = |cd| and |ad| = |bc| Construction: Diagonal |ac| Proof: b 4 c 2 ∠1 = ∠4 Alternate angles |ac| = |ac| Given ∠2 = ∠3 Alternate angles ASA ⇒ ∆ abc = ∆ acd a 3 ⇒ 1 d |ab| = |cd| and |ad| = |bc| Q.E.D. Theorem 3: If three parallel lines make intercepts of equal length on a transversal, then they will also make equal length on any other transversal. d b1 2 c 3 4 Given: Diagram as shown with |ab| = |bc| To Prove: |db| = |be| Construction: Another transversal through b. Proof: ∠1 = ∠2 Verticially opposite |ab| = |bc| Given ∠3 = ∠4 Alternate angles a ASA ⇒ ∆ dab = ∆ bec e ⇒ |db| = |be| Q.E.D. Theorem 4: A line which is parallel to one side-line of a triangle, and cuts a second side, will cut the third side in the same proportion as the second. Given: Diagram as shown with line |xy| parallel to base. To Prove: |xb| a b y c m equal parts n equal parts x |ax | = |ay| |yc| Construction: Draw in division lines Proof: |ax | is divided in m equal parts ⇒ |ay| is also |xb| is divided in n equal parts ⇒ |yc| is also |ax | |xb| = Q.E.D. |ay| |yc| Theorem 2 Theorem 5: If the three angles of one triangle have degree-measures equal, respectively, to the degree-measures of the angles of a second triangle, then the lengths of the corresponding sides of the two triangles are proportional. Given: Two Triangles with equal angles To Prove: |ab| = |de| x 4 b |df | = |bc| |ef | a d Construction: Map ∆ def onto ∆ axy, draw in [xy] 2 2 Proof: ∆ def mapped onto ∆ axy ∠1 = ∠4 5 y e 1 3 ⇒ f ⇒ 1 |ac| 3 [xy] is parallel to [bc] |ab| |ax| |ab| c |de| = = Q.E.D. |ac| Theorem 3 |ay| |ac| |df | Similarly = |bc| |ef | Theorem 6: In a right-angled triangle, the square of the length of the side opposite to the right-angle is equal to the sum of the squares of the lengths of the other two sides. (Pythagoras) b a a c c b c a Given: Triangle abc To Prove: a 2 + b2 = c 2 Construction: Three right angled triangles as shown b Proof: Area of large sq. = area of small sq. + 4(area ∆) (a + b)2 = c2 + 4(½ab) a2 + 2ab +b2 = c2 + 2ab c b a a 2 + b2 = c 2 Q.E.D. Theorem 7: If the square of the length of one side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle has a right-angle and this is opposite the longest side. (Converse of Pythagoras’ Theorem) Triangle abc with |ac|2 = |ab|2 + |bc|2 Given: Construction: Triangle def with |de| = |ab| |ef | = |bc| & d a b 1 c e 2 ∠2 = 900 To Prove: To prove ∠1 = 900 Proof: |ac|2 = |ab|2 + |bc|2 given |ac|2 = |de|2 + |ef|2 from construction |ac|2 = |df|2 |df|2 = |de|2 + |ef|2 f ⇒ |ac| = |df| SSS ⇒ ∆ abc = ∆ def ∠1 = 900 Q.E.D. Theorem 8: The products of the lengths of the sides of a triangle by the corresponding altitudes are equal. Given: Triangle abc To Prove: |bc||ad| = |ac||be| Construction: Altitudes [be] and [ad] Compare ∆ bce to ∆ acd Proof: 2 a b 5 2 b 3 4 d c ∠1 = ∠1 e ∠2 = ∠4 ⇒ 1 5 1 3 a e ∠3 = ∠5 Similar triangles c |bc| |ac| ⇒ = |be| |ad| |bc||ad| = |ac||be| Q.E.D. d 4 1 c Theorem 9: If lengths of two sides of a triangle are unequal, then the degree-measures of the angles opposite to them are unequal, with the greater angle opposite to the longer side. Given: Triangle abc To Prove: ∠abc > ∠acb Construction: Draw [bd] such that |ab| = |ad| ∠1 = ∠2 Proof: a 2 b 1 3 ⇒ ∠1 + ∠3 > ∠2 But ∠2 > ∠4 ⇒ ∠1 + ∠3 > ∠4 ⇒ ∠abc > ∠acb As |ab|= |ad| and the greater angle is d opposite the longer side. 4 c Q.E.D Theorem 10: The sum of the lengths of any two sides of a triangle is greater than that of the third side. Given: Triangle abc To Prove: |ba| + |ac| > |bc| Construction: Draw ∆ acd such that |ad| = |ac| In ∆ acd |ad| = |ac| Proof: ∠1 = ∠2 d 2 a b ⇒ ∠1 + ∠3 > ∠2 ⇒ |bd| > |bc| But |bd| = |ba| + |ad| 1 3 c |bd| = |ba| + |ac| ⇒ |ba| + |ac| > |bc| Q.E.D. as |ad| = |bc|