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Given two points (x 1 ,y 1 ) and (x 2 ,y 2 ). Transversals. Equal angles: corresponding (A and E, B and F, C and G, D and H) alternate interior (C and F, D and E) alternate exterior (A and H, B and G). a. b. c. d. e. f. g. h. Same side interior - angles add up to 180

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## Given two points (x 1 ,y 1 ) and (x 2 ,y 2 )

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**Transversals**Equal angles: • corresponding • (A and E, B and F, C and G, D and H) • alternate interior • (C and F, D and E) • alternate exterior • (A and H, B and G) a b c d e f g h Same side interior - angles add up to 180 (C and E, D and F)**Circles**- set of all points in a plane equidistant from a center d = 2r C = dπ A = πr2 secant chord radius O diameter tangent**Theorems**• the perpendicular from the center of a circle to a chord bisects the chord(AB bisects CD) • the segment from the center of the circle to the midpoint of a chord is perpendicular to it (AB is perpendicular to CD) • the perpendicular bisector of a chord passes through the center (the line bisecting CD will pass A) C B A D**chords equidistant from the center are congruent (AB and CD)**• congruent chords have congruent arcs (AB and CD) A B C D**in a quadrilateral inscribed within a circle ("cyclic**quadrilateral"), the opposite angles are supplementary (A and C, B and D) • parallel secants have congruent arcs (EF and GH) G B E A D H C F**Secant and tangent**(AB)(AC) = (DA)2 Power Theorems • Two tangents AC = BC C B A D A C B**Two chords**(AB)(BC) = (DB)(BE) D A • Two secants (AB)(AC) = (AD)(AE) B C D A E B E C**the measure of the angle of two chords intersecting within a**circle is equal to half the sum of their intercepted arcs • ½ (a + b) = x a x b**x**x a a b b b a x**Inscribed Angle**- vertex is on the circle x = a/2 Central Angle - vertex is at the center • x = a**Sectors**- region enclosed by two radii and the intercepted arc • A = x/360 πr2 • Segment of a circle - region between a cord and intercepted arc - A = (Asector ) - (Atriangle)**Polygons**Convex polygons - all diagonals lie entirely inside the polygon Regular polygons - equilateral and equiangular**Diagonals - segments connecting nonconsecutive vertices of**the polygon • number of diagonals = • Sum of the measures of interior angles: (n-2)180 Measure of each interior angle: Sum of the measures of exterior angles: 360 Measure of each exterior angle: 360/n**Rectangles**P = 2 (l + w) A = lw d2 = (l2 + w2) (look pythagorean theorem again) Rectangular Box: V = lwh S = 2lw + 2lh + 2wh d = l2 + w2 + h2 Squares P = 4s A = s2 = d2/2 d2 = s2 Cube: V = s2 S = 6s2 d2 = 3s2**Trapezoid**A = ½ h(b1+b2) midsegment = (b1+b2)/2 Kite A = ½ (d1d2) * diagonals are perpendicular Other cool shapes Parallelogram A = bh * consecutive angles are supplementary Rhombus A = bh = ½ (d1d2) * equilateral parallelogram**Cylinder**V = πr2h S = 2πr2+ 2πrh Pyramid V = (s2h)/3 Cone V = 1/3 πr2h Sphere V = 4/3 πr3 S = 4πr2**Apothem**= in a regular polygon, is the perpendicular distance from the center of each of the sides (it's like the radius of a polygon) • A = ½ (apothem)(perimeter)**Triangles**- angles add up to 180 Sides: c < a + b (triangle inequality theorem) P = a + b + c A = ½ (bh) • Equilateral Triangles • A = s2 sqrt ¾ • P = 3s**Right Triangles**- Pythagorean Theorem: c2 = a2 + b2 Special Right Triangles x 30° 45° 90° x√3 2x √2 x x 45° 60° 90° x**Similarity**- if the corresponding angles are equal, and the corresponding sides are proportional - AAA, SSS, and SAS Similarity Theorems**Basic Proportionality Theorem**• if a line/segment parallel to one side of a triangle intersects the other two sides in distinct points, then cuts off segments proportional to those sides • DE ║ BC; AB/AD = AC/AE and BD/AD = CE/AE A D E C B**Midline Theorem**- The segment connecting the midpoints of the two sides of a triangle is equal to half the side it is parallel to. - MN = ½ BC A M N C B**- The altitude to the hypotenuse of a right triangle divides**it into two similar right triangles (both are similar to the original triangle) A Triangles ABC, ACD, and CBD are similar B D C**Congruence**- triangles are congruent if the corresponding sides and angles are congruent - SAS, ASA, SSS, SAA postulates - Hypotenuse-Leg theorem**Exterior Angle Theorem**- the measure of an exterior angle is equal to the sum of the measures of its two remote interior angles 1 = a + b b 1 a**Side-Angle Inequality Theorem**- If the angles aren't congruent, the sides aren't either (the converse is also trueee) Isosceles Triangle Theorem - If the two sides are congruent, the two angles opposite them are also congruent (the converse is also trueee)**Degree and Radian Measures**2 π (radians) = 1 revolution = 360° So the Quadrantal Angles are at 0° 90° (½π, ¼ of a revolution) 180° (π, ½ of a revolution) 270° (3/2 π, 3/4 of a revolution)**Trigonometric Ratios**sin = opposite/hypotenuse cos = adjacent/hypotenuse tan = opposite/adjacent • csc= hypotenuse/opposite • sec = hypotenuse/adjacent • cot = adjacent/opposite Cofunctions Sine and Cosine [sinA = cos (90-A) andcosA = sin (90-A)] Secant and Cosecant Tangent and Cotangent**Trigonometric Identities**• sin (–θ) = –sin θ • cos (–θ) = cosθ • tan (–θ) = –tan θ • csc (–θ) = –cscθ • sec (–θ) = secθ • cot (–θ) = –cotθ • ** Basically everything becomes negative except cos and sec sin2θ + cos2θ = 1 tan2θ + 1 = sec2θ cot2θ + 1 = csc2θ • sin θ = 1/csc θ • cos θ = 1/sec θ • tan θ = 1/cot θ • tan θ = sin θ / cosθ • cot θ = cosθ /sin θ

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