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Math 531, Exam 1 Information Math 531, Exam 1 Information. 9/21/11, LC 310, 9:05 - 9:55. Exam 1 will be based on: • Sections 1A - 1F. • The corresponding assigned homework problems (see http://www.math.sc.edu/∼boylan/SCCourses/531Fa11/531.html) At minimum, you need to understand how to do the homework problems. Topic List (not necessarily comprehensive): You will need to know: theorems, results, and definitions from class. x1B: Congruent triangles. Definition: 4ABC ∼= 4RST if and only if 1. 3 pairs of corresponding sides are equal: (a) AB = RS, (b) BC = ST , (c) CA = TR and 2. 3 pairs of corresponding angles are equal: (a) \A = \R, (b) \B = \S, (c) \C = \T . Note: 4ABC ∼= 4RST does not imply that 4ABC ∼= 4T SR. Notation: S = side; A = angle: Example: For triangles T1 and T2, “SAS” (“side-angle-side”) means that the triangles have two pairs of equal sides with the angle between the sides equal. Table 1: Which conditions guarantee a congruence? condition congruence? (Y/N) SSS Y SSA N SAS Y ASS N SAA Y ASA Y AAS Y AAA N Definition: Let 4ABC be a right triangle. Its largest side is the hypotenuse. The other two sides are its arms. Fact: Suppose that 4ABC and 4DEF are right triangles with equal hypotenuses and one pair of equal arms. Then the triangles are congruent. I.e., we have “HA” =) congruence. (Thm. 1.3). Definitions: Consider 4ABC. 1. median. Let X = midpt(BC). Then AX = med(A), the median from A to side BC. 2. altitude. Let Y be the point on side BC for which AY is perpendicular to BC. Then AY = alt(A), the altitude from A to BC. 3. angle bisector. Let Z be the point on side BC such that \BAZ = \CAZ. Then AZ = bis(\A), the bisector of \A. Facts on isosceles triangles: (1) A triangle 4ABC is isosceles with base BC () its base angles are equal: \B = \C. Note: =) is Theorem 1.2 (pons asinorum); the converse is homework problem 1B.1. (2a) Suppose: 4ABC is isosceles with base BC. Then we have med(A) = alt(A) = bis(\A). Note: This conclusion only applies to the angle opposite the base. (Theorem 1.2) (2b) Consider 4ABC. Suppose that any two of the following are equal: fmed(A); alt(A); bis(\A)g. Then the triangle is isosceles. (Homework problems 1B.2, 1B.3, and Problem 1.13.) 2 x1C: Angles and parallel lines. Vocabulary: 1. Vertical angles. 2. Exterior angles. 3. Remote interior angles with respect to an exterior angle. 4. Alternate interior angles. 5. Corresponding angles. Facts: 1. Vertical angles are equal. 2. Corresponding angles are equal 3. Alternating interior angles are equal. 4. An exterior angle is larger than either of its two corresponding remote interior angles. Moreover, an exterior angle is the sum of its two corresponding remote interior angles. (Thm. 1.4). Definition: An interior angle of an n-gon is a reflex angle () it is > 180◦. Theorem: The sum of the interior angles of an n-gon is 180(n − 2) degrees. x1D: Parallelograms. Definition: A quadrilateral is a 1. parallelogram () its opposite sides are parallel. 2. rhombus () all 4 sides are equal. 3. rectangle () all 4 angles are equal (hence, 90◦). 4. square () it is a rhombus and a rectangle. Fact: A quadrilateral ABCD is a parallelogram if and only if any one of the following is true: 1. Its opposite sides are equal: AB = CD, AD = BC. (Theorems 1.6 and 1.7) 2. Its opposite angles are equal: \A = \C, \B = \D. (Homework 1D.7, 1D.9) 3. It has one pair of opposite sides which are parallel and have equal length. (Example: AB = CD and AB k CD.) (Theorem 1.8, homework problem 1D.2) 4. Its diagonals, AC and BD, bisect each other. (Theorem 1.9) 3 Further results: 1. Adjacent interior angles of a parallelogram are supplements (add to 180◦). 2. Suppose that parallelogram ABCD has an interior right angle. Then it is a rectangle. 3. Let BC be a given line segment. Then the set of all points equidistant from B and C, L = fP : BP = CP g; is the perpendicular bisector of BC. (Theorem 1.10) 4. Parallelogram ABCD is a rhombus () its diagonals are perpendicular. (Corollary 1.11 and homework 1D.4). x1E: Area. Basic facts and definitions. 1. Parallelogram: Area = base × height. 2. Triangle: Area = (1=2) Base × height. opposite 3. sine: In a right triangle, the sine of an angle θ is sin θ = . It is well - defined hypotenuse (only depends on θ, not on side lengths of the triangle). Law of Sines: Consider 4ABC. We have a b c = = : sin A sin B sin C Further triangle area formulas. Consider 4ABC. Its area is expressible as: 1 1 1 1. (SAS): Area = 2 ab sin C = 2 ac sin B = 2 bc sin C. a+b+c 2. (SSS): Heron’s Formula: Let s = 2 be the semiperimeter of 4ABC. Then we have Area = s(s − a)(s − b)(s − c): 3. There are other formulas corresponding to the fact that AAS, SAA, and ASA are valid triangle congruence criteria. Homework 1E.3 addresses ASA. È BX AB Auxiliary result: Consider 4ABC. Suppose that bis( A) = AX. Then we have = . \ XC AC 4 x1F: Circles and arcs. Vocabulary: 1. Arc. 2. Central angle corresponding to an arc. 3. Chord. 4. Secant. 5. Tangent. 6. Circumcircle of 4ABC. 7. Angle inscribed in a circle. 8. Polygon inscribed in a circle. Notation: The symbol $ means “equal in degrees or radians”. Therefore, AB $ θ means that the arc AB has corresponding central angle θ. Note that we can have AB $ CD (corresponding central angles equal in degrees) but AB 6= CD (arcs are not the same length). Theorem: Let A, B, and C be non-collinear points. There exists a uniqueø circle passing through theseø points. (3 non-collinear points uniquely determine a circle.) (Theoremø 1.15)ø ø ø Facts: The following are basic relations between arcs and angles formed different ways. 1. Suppose that inscribed angle \P subtends arc AB. Then we have \P $ (1=2)AB. (Theorem 1.16) 2. Suppose that two secants meet at a point X outside of the circle, and suppose that \X subtends large arc AB and small arc RP . Then we have \X $ (1=2)(AB − RP ). (Theorem 1.18) ø ø 3. Suppose that two chords meet at a point X inside of the circle, and suppose that \X subtends the two arcs AB and RP . Then we have \X $ (1=2)(AB + RP ). (Theorem 1.19) ø ø ø ø 4. Suppose that T is the line tangent to the circle at point P , and suppose that QP is a chord. Then we have \QP T $ (1=2)QP . (Theorem 1.23) ø ø ø ø Facts on tangents: 1. Let P be a point on circle C. There existsø a unique line tangent to C at P . 2. The line tangent to C at point P is perpendicular to the radius through P . 5 Further results. 1. A quadrilateral inscribed in a circle has opposite angles supplementary. (Theorem 1.17) ◦ 2. Suppose that C is the circumcircle of 4ABC. Then we have \C = 90 () AB is the diameter of C. (Theorem 1.22) 3. Suppose that AB and CD are chords on two circles with equal radii. Then the chords are equal (AB = CD) if and only if AB $ CD. (Homework 1F.1) ø ø 6.
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