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Proportions: a Ratio Is the Quotient of Two Numbers. for Example, 2 3 Is A

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Proportions: a Ratio Is the Quotient of Two Numbers. for Example, 2 3 Is A Proportions: 2 A ratio is the quotient of two numbers. For example, is a ratio of 2 and 3. 3 An equality of two ratios is a proportion. For example, 3 15 = is a proportion. 7 45 If two sets of numbers (none of which is 0) are in a proportion, then we say that the numbers are proportional. a c Example, if = , then a and b are proportional to c and d. b d a c Remember from algebra that, if = , then ad = bc. b d The following proportions are equivalent: a c = b d a b = c d b d = a c c d = a b Theorem: Three or more parallel lines divide trasversals into proportion. l a c m b d n In the picture above, lines l, m, and n are parallel to each other. This means that they divide the two transversals into proportion. In other words, a c = b d a b = c d b d = a c c d = a b Theorem: In any triangle, a line that is paralle to one of the sides divides the other two sides in a proportion. E a c F G b d A B In picture above, FG is parallel to AB, so it divides the other two sides of 4AEB, EA and EB into proportions. In particular, a c = b d b d = a c a b = c d c d = a b Theorem: The bisector of an angle in a triangle divides the opposite side into segments that are proportional to the adjacent side. C b a A B c D d In 4ACB above, CD is the bisector of \ACB. We have: a b = d c d c = a b d a = c b c b = d a Similar Triangles: Two triangles are said to be similar to each other if they have the same shape (not necessarily the same size). If two triangles are similar, then all three of their angles are congruent to each other, and their corresponding sides are in proportion. This means that the ratio of their corresponding sides are equal to each other. We use the ∼ symbol to mean similar to. So we write 4ABC ∼ 4DEF if triangle ABC is similar triangle DEF . 0 0 0 ∼ 0 ∼ In the picture below, 4ABC ∼ 4A B C . This means that \A = \A , \B = 0 ∼ 0 \B , and \C = \C . Further more, their corresponding sides are in proportion, so AB A0B0 AB A0B0 BC B0C0 = = = AC A0C0 BC B0C0 AC A0C0 C0 C A B A0 B0 Theorem: If 4ABC ∼ 4DEF , and 4DEF ∼ GHI, then 4ABC ∼ 4GHI. In other words, similarity is transitive. AA (Angle-Angle) Similarity Theorem: If two angles of a triangle is congruent to two angles of another triangle, then the two triangles are similar. C0 C A B A0 B0 ∼ 0 ∼ 0 0 0 0 In picture above, \A = \A , \B = \B , therefore, 4ABC ∼ 4A B C SAS (Side Angle Side) Similarity Theorem: If an angle is congruent to an angle of another triangle the the sides that make up the angles are in proportion, then the two triangles are similar. C0 C b0 a0 b a A B A0 B0 ∼ 0 0 0 0 0 In picture above, if \C = \C , and AC, BC are in proportion with A C , B C , a a0 meaning that = , then 4ABC ∼ 4A0B0C0 b b0 SSS (Side-Side-Side) Similarity: If all three sides of a triangle are in propor- tion with three sides of another triangle, then the two triangles are similar. C0 C b0 a0 b a A B A0 B0 c c0 In picture above, if the three sides of 4ABC is proportional to the three sides of 4A0B0C0, meaning that: a b c = = , a0 b0 c0 then 4ABC ∼ 4A0B0C0 Theorem: A line parallel to one of the sides of a triangle divides the triangle into a smaller triangle similar to the original one. E F G A B In picture above, if FGjjAB, then 4F EG ∼ 4AEB Theorem: The altitude of a right triangle (through the right angle) divides the triangle into two similar right triangles that are also similar to the original right triangle. C A B D In picture above, 4ABC is a right triangle with \ACB being the right angle. CD is the altitude throught C perpendicular to AB. We will prove that 4ACB ∼ 4ADC ∼ 4CDB Proof: Statements Reasons 1. CD is altitude to 4ACB 1. given 2. AB ? CD 2. Def. of Altitude 3. \CDB and \CDA are right \'s 3. Def. of ? lines ◦ 4. m\ACD + m\BCD = 90 4. Angle Addition Postulate 5. \ACD complementary to \BCD 5. Def. of Comp. Angles ◦ 6. m\CAD + m\ACD = 90 6. Sum of Interior Angles of 4 7. \ACD complementary to \CAD 7. Def. of Comp. Angles ∼ ∼ 8. \BCD = \CAD 8. \'s Comp. to same \ are = ◦ 9. m\CBD + m\BCD = 90 9. Sum of Interior Angles of 4 10. \CBD complementary to \BCD 10. Def. of Comp. Angles ∼ ∼ 11. \CBD = \ACD 11. \'s Comp. to same \ are = 12. 4ADC ∼ 4CDB 12. AA ∼ 13. 4ACB ∼ 4ADC 13. AA ∼ The previous theorem about the altitude of a right triangle dividing the right triangle into three similar triangles allows us to prove one of the most important theorem in geometry (and mathematics in general): Pythagorean Theorem: In any right triangle, the sum of the square of the length of the two legs is equal to the square of the length of the hypotenuse. C b h a d e A B D c D D e h h d A C C B b a To prove the pythagorean theorem, note from the previous theorem that 4ACB ∼ 4ADC ∼ 4CDB. Therefore, their corresponding sides are in proportion. In particular, b d a e = ; = c b c a The first equation gives us: b d b2 = ) d = c b c The second equation gives us: a e a2 = ) e = c a c Notice that e + d = c, therefore, we have: a2 b2 a2 + b2 e + d = + = c ) = c ) a2 + b2 = c2 c c c Example: 5 7 h 2 p c 6 p p In first triangle, 52 + c2 = 72 ) 25 + c2 = 49 ) c2 = 24 ) c = 24 = 2 6 p 2 p In second triangle, 6 + (2)2 = h2 ) h2 = 6 + 4 = 10 ) h = 10 Special Triangles: A right triangle where one of the acute angle is 30◦ and the other acute angle is 60◦ is called a 30-60-90 triangle. ◦ Theorem: In a 30-60-90 triangle, if the side oppositep the 30 angle has length of a, then the side opposite the 60◦ angle has length of 3a, and the hypotenuse has length 2a. B 60◦ 2a a 30◦ A p C 3a ◦ In 4ABC, BC is the side opposite the 30 angle \A, if length of pBC = a, then the side opposite the 60◦ angle, side AC, will have a length of 3a, and the hypotenuse has length 2a. We can prove the 30◦ − 60◦ − 90◦ triangle by drawing the altitude CD to the equilateral triangle ABC as above. Let b be the length of each of its side. Notice that the altitude CD divides the eqilateral triangle into two congruent triangles (why?). Each of the triangle is a 30 − 60 − 90 triangle and the length of AD is half of b. Using Pythagorean theorem, we see that ! b 2 b2 b2 + h2 = b2 ) + h2 = b2 ) h2 = b2 − ) 2 4 4 p 4b2 b2 3b2 3b bp h2 = − ) h2 = ) h = = 3 4 4 4 2 2 C 30◦ b b h ◦ 60 b=2 D b=2 A B b A 45-45-90 triangle is an isoceles right triangle. In an isoceles right triangle, each acute angle is the same, therefore, each acute angle is 45◦. ◦ ◦ ◦ Theorem: In a 45 − 45 − 90 triangle, if the length of one of the leg is ap, then the length of the other side is also a, and the length of the hypotenuse is 2a B 45◦ p 2a a 45◦ A a C The 45◦ − 45◦ − 90◦ triangle can be proved just using the Pythagorean theorem. In 4ABC, if one of the leg has length a, the the Isoceles triangle theorem tells us the other leg must also has length a since the base angles are congruent. Using the Pythagorean theorem, the length of the hypotenuse, h, is: p h2 = a2 + a2 = 2a2 ) h = 2a If two chords intersect inside a circle, the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other. B C E A D In the circle above, AC, BD are chords intersect at E, the theorem tells us that: AE · EC = BE · ED If two secant lines intersect outside a circle, the product of the length of one of the secants with the length outside the circle is equal to the product of the length of the other secant with the length outside the circle. D C A E B In circle above, secants AC, BC intersect the circle at points D and E, respec- tively, and we have: CA · CD = CB · CE If a tangent and a secant to a circle intersect outside the circle, the square of the length of the tangent is equal to the product of the length of the secant times the length of the secant outside the circle. B C D A In circle above, tangent BC intersects AC outside the circle at C, we have: CB · CB = CA · CD.
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