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Published byJeffery Johns Modified over 9 years ago
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Parallelism Triangles Quadrilaterals
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Key Terms skew lines: non-coplanar lines that do not intersect (lines AB and EF are skew lines) parallel lines: non-intersecting coplanar lines (lines AD and EF are parallel lines) A B C D EF
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transversal: line that intersects two coplanar lines (line AB is a transversal) alternate interior angles: angles that lie inside two lines and on opposite sides of the transversal ( r and s are alternate interior angles) corresponding angles: if r and s are alternate interior angles and q is a vertical angle to r, then q and s are corresponding angles. q r s A B
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Theorems AIP Theorem If you are given two lines intersected by a transversal, and a pair of alternate interior angles are congruent, then the lines are parallel. Restatement: Given line AB and line CD cut by transversal EF. If x y, then line AB is parallel to line CD. x y AB CD E F
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Given: line segment GH and line segment JK bisect each other at F. Prove: line segment GK and line segment JH are parallel. 1. GH and JK bisect at F 2. GF = FH and JF = FK 3. JFH KFG 4. ∆JFH ∆KFG 5. HJF GKF 6. JH is parallel to GK G HJ K F 1. Given 2. Def. of bisector 3. VAT 4. SAS 5. CPCTC 6. AIP StatementsReasons
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PCA Corollary Corresponding angles are congruent if you are given two parallel lines cut by a transversal. Restatement: v w if line AB and line CD are parallel and are cut by transversal EF. AB C D E F v w
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Given: the figure with CDE A and line LF line AB. Prove: line LF line DE. 1. CDE A, LF AB 2. DE is parallel to AB 3. GFA LGD 4. GFA is a R.A. 5. m GFA = 90˚ 6. m LGD = 90˚ 7. LGD = R.A. 8. LF DE 1. Given 2. CAP 3. PCA 4. Def. of perp. 5. Def. of R.A. 6. Def. of congruence 7. Def. of R.A. 8. Def. of perp. StatementsReasons AB C D E G L H
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Key Terms concurrent lines: two or more lines that all share a common point (lines AB, CD, and EF are concurrent.) point of concurrency: the common point shared by concurrent lines. (point G is the point of concurrency.) A B C D E F G
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Theorems Theorem 9-13 The measures of all the angles in a triangle add up to 180. Restatement: Given ABC. m A + m B +m C = 180. A B C
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Given: ABC, BA AC and m B = 65. Prove: m C = 155. Statements 1. BA AC, m B = 65 2. A is a R.A. 3. m A = 90 4. m A + m B +m C = 180 5. 90 + 65 + m C = 180 6. m C = 155 Reasons 1. Given 2. Def. of perp. 3. Def. of R.A. 4. m s in = 180 5. Sub. 6. SPE A B C 65˚
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Theorem 9-28 If one side of a right triangle is half the length of the hypotenuse, then the measure of the opposite angle is 30. Restatement: Given right triangle ABC. If AB = 1/2BC, then m BCA is 30. A B C
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Given: DEF is a right triangle. D = 90 and DE = 1/2EF. Prove: m E = 60 Statements 1. DEF is a R.T. DE = 1/2EF 2. m F = 30 3. m D = 90 4. m D + m E + m F = 180 5. 90 + m E + 30 = 180 6. m E = 60 Reasons 1. Given 2. opp. side 1/2 as long as hyp. = 30 3. Def. of R.T. 4. s add up to 180 5. Sub. 6. SPE D E F
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Key Terms diagonal: a line segment connecting two nonconsecutive angles in a quadrilateral. (segment AC is a diagonal) parallelogram: quadrilateral that has opposite parallel lines. (ABCD is a parallelogram) trapezoid: quadrilateral that has one pair of opposite parallel lines and one pair of nonparallel lines. (JKLM is a trapezoid.) A BC D J KL M
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rhombus: parallelogram that has 4 congruent sides. (QRST is a rhombus) rectangle: parallelogram with 4 right angles. (ABCD is a rectangle) square: rectangle that has 4 congruent sides. (FGHJ is a square) Q R S T A BC D F G H J
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Theorems Theorem 9-21 If the diagonals in a quadrilateral bisect each other, that quadrilateral is a parallelogram. Restatement: Given ABCD. If AC and BD bisect each other at E, then ABCD is a parallelogram. A B D C E
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Given: WXYZ WT = TY and XT = TZ Prove: WXYZ is a parallelogram. Statements 1. WT = TY, XT = TZ 2. WY and XZ bisect each other. 3. WXYZ is a parallelogram. Reasons 1. Given 2. Def. of bisectors 3. If diagonals bisect each other, = parallelogram W XY Z T
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Theorem 9-24 A rhombus’ diagonals are perpendicular to each other. Restatement: Given ABCD is a rhombus, then AC BD. A B C D
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Given: FGHK is a rhombus. Prove: GJF GJH Statements 1. FGHK is a rhombus 2. GK FH 3. GJK and GJH are R.A. 4. GJK GJH Reasons 1. Given 2. Diagonals are in rhombus 3. Def. of perp. 4. R.A. F G H K J
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