Transcript Matt
Period 2 Chapter 9 Summary Project 12/17 Starring: & Triangles Quadrilaterals By Matthew Donoghue Para e ism Corresponding Angles: Definitions Alternate Interior Angles: Two angles that are formed by two lines cut by a transversal. Each angle is located on the same side of the transversal and one is interior, and the other is exterior. Two angles that are formed by two lines cut by a transversal. Each angle is located on an opposite side of the transversal and neither share a common ray. Skew Lines: Parallel Lines: Two lines that are coplanar and never intersect. Two lines that are non-coplanar and never intersect. Transversal: A line that intersects two lines at different points. Theorem 9-5 The AIP Theorem If two lines cut by a transversal form two congruent alternate interior angles, then the lines are parallel. = Restatement. Given: Lines L1 and L2 cut by a transversal T. a b, then L1 L2 Statement Reason 1. AC = BD, AB = CD 1. Given Parallelism Problem #1. 2. CB = CB 2. RPE Use Diagram above right. 3. ∆ABC ∆DCB 3. SSS = Prove L1 L2. ACB 5. L1 L2 = Given: AC = BD, AB = CD. Lines L1 and L2 are cut by a transversal T. 4. CBD 4. CPCTC 5. AIP Theorem 9-30 It has no name, so don’t ask for one. Just 9-30. Say bye-bye to parallelism. = = Reason 1. L1 L2 L3, L4 L5 L6, AB = BC. 2. GH = HI. 3. FE = DE. 1. Given == Restatement Given: L1 L2 L3. All three lines are intersected by transversals T1 and T2; AB = BC. Then EF = FG. Statement == If two congruent segments are cut by three parallel lines, then any other transversals along the lines are also cut in to equal segments. Parallelism Problem #2 Prove: EF = DE. = = = = Use figure below right. Given: L1 L2 L3, L4 L5 L6. L1, L2, & L3 are cut by transversals T1 & T3. L4, L5, & L6 are cut by transversals T1& T2. AB = BC. 2. 9-30 3. 9-30 Trangle Definitions Theorems Statement Reason Isosceles Triangle 1. MAT and ROX 1. Given are right triangles; MT = 20; R & T = 30˚ 2. MA = 10 2. 30-60-90 3. MA = RX 3. Given 4. RX = 10 5. OX = 5 Theorem 9-27 The 30-60-90 Triangle Theorem In a right triangle, if the smallest angle measures 30˚, then the shortest side, which is opposite the 30˚ angle, is 1/2 the length of the hypotenuse. Restatement 4. Substitution Given: ABC is a right triangle; A has a measure of 30˚; D Right Triangle is the midpoint of AB. Then BC = 1/2 AB. 5. 30-60-90 Triangles Problem #1 An isosceles triangle has a pair of congruent angles and sides. The two congruent sides will always be opposite the two congruent angles, and vice versa. Use the. figure on the left. Given: MAT and ROX are right triangles; MA = RX; R & T = 30˚; MT = 20. A right triangle has one right and two acute angles. P.S. Right triangles can be isosceles as well. Corollary 9-13.3 Un-named. The exterior angle of any triangle, has the same degrees as the two remote interior angles added together. Restatement Given: XYZ with angles A, B, C, and E. E is an exterior angle adjacent to angle C. Then the sum of A & B = E. Triangles Problem #2 Use figure to below. Given: RSU ( R + RSQ). Prove: RSQ is isosceles. Statement 1. RSU ( + RSQ). 2. TQR ( + RSQ). Reason R R 1. Given 2. Corollary 913.3 3. TQR RSU. 3. TPE 4. TQR is supp. 4. Supp. Pos. to SQR. 5. USR is supp. 5. Supp. Pos. to QSR. 6. QSR SQR. 6. Supp. Theorem. 7. RSQ is isosc. 7. Def of Isosc. Quadrilateral Square Definitions Rectangle Quadrilateral Any 2 dimensional figure with exactly 4 sides. Any parallelogram with 4 right angles. A parallelogram with 4 right angles and 4 congruent sides. Rhombus Parallelogram Any quadrilateral with every pair of opposite sides being parallel. Trapezoid Any quadrilateral with only one pair of opposite sides being parallel. Any parallelogram with 4 congruent sides. Theorem 9-16 Un-named In a parallelogram, the opposite angles are congruent. Restatement Given: Parallelogram •ABCD, then A C & B D. Quadrilaterals Problem #1 Use the figure to the right. Given: Parallelogram •ABCD with FB = HD & BE = GD. Statement Reason Prove: EF = GH. 1.•ABCD is a parallelogram; BE = GD; FB = HD 2. B D. 3. FBE HDG 1. Given 4. EF = GH 4. CPCTC 2. 9-16 3. SAS Theorem 9-25 Un-named If a quadrilaterals diagonals are perpendicular to each other and bisect each other, then the quadrilateral is a rhombus. Restatement Given:•ABCD, with AC BD, and AC and BD bisecting each other, then it is a rhombus. Quadrilateral Problem #2 Use the figure on the right. Given:•ABCD , with AC BD, and AC and BD bisecting each other. Prove: All sides are equal. Statement 1.•ABCD , with AC BD, and AC and BD bisecting each other. 2.•ABCD is a rhombus. 3. All sides are equal. Reason 1. Given 2. 9-25 3. Def of rhombus