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Exploring Triangle and Quadrilateral Properties through Parallelism

This project summary delves into the fundamental concepts of parallelism, triangles, and quadrilaterals as explored by Matthew Donoghue. Key definitions such as corresponding angles, alternate interior angles, and properties of skew lines are discussed. Theorems, including the Alternate Interior Angles Theorem and the criteria for identifying congruent triangles, are emphasized. Additionally, the characteristics of various quadrilaterals like rectangles, rhombuses, and parallelograms are examined, providing a comprehensive overview of their geometric properties and relationships.

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Exploring Triangle and Quadrilateral Properties through Parallelism

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  1. Period 2 12/17 Chapter 9 Summary Project Starring: Parallelism & Triangles Quadrilaterals By Matthew Donoghue

  2. P a r a e i s m Corresponding Angles: Definitions 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. Alternate Interior Angles: 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: Transversal: Parallel Lines: A line that intersects two lines at different points. Two lines that are non-coplanar and never intersect. Two lines that are coplanar and never intersect.

  3. Theorem 9-5 T The AIP Theorem A L2 C If two lines cut by a transversal form two congruent alternate interior angles, then the lines are parallel. a b L1 B D Restatement. Given: Lines L1 and L2 cut by a transversal T. a  b, then L1 L2 < < = Statement Reason 1. Given 1. AC = BD, AB = CD Parallelism Problem #1. 2. RPE 2. CB = CB 3. ∆ABC  ∆DCB 3. SSS Use Diagram above right. Given: AC = BD, AB = CD. Lines L1 and L2 are cut by a transversal T. 4. ACB  CBD 4. CPCTC < < = 5. AIP 5. L1 L2 = Prove L1 L2.

  4. Theorem 9-30 Say bye-bye to parallelism. It has no name, so don’t ask for one. Just 9-30. T1 T2 L1 A E L2 B F If two congruent segments are cut by three parallel lines, then any other transversals along the lines are also cut in to equal segments. L3 C G T1 Statement Reason = = 1. L1 L2 L3, 1. Given = = L4 L5 L6, Restatement Given: L1 L2 L3. All three lines are intersected by transversals T1 and T2; AB =BC. Then EF = FG. AB = BC. T2 = = 2. 9-30 2. GH = HI. D E F 3. 9-30 3. FE = DE. L1 A I Parallelism Problem #2 L2 B H 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. L3 C G = = = = L6 T3 L5 Prove: EF = DE. L4

  5. Trangle Theorem 9-27 Definitions Theorems The 30-60-90 Triangle Theorem Statement Reason B Isosceles Triangle 1. MAT and ROX are right triangles; MT = 20; R & T = 30˚ 1. Given 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. D 30˚ A C < < 2. MA = 10 2. 30-60-90 3. MA = RX 3. Given Restatement Given: ABC is a right triangle; A has a measure of 30˚; D is the midpoint of AB. Then BC = 1/2 AB. 4. RX = 10 4. Substitution < Right Triangle 5. OX = 5 5. 30-60-90 Triangles Problem #1 Use the. figure on the left. Given: MAT and ROX are right triangles; MA = RX; R & T = 30˚; MT = 20. 5 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. O R M < < 30˚ 10 10 A right triangle has one right and two acute angles. P.S. Right triangles can be isosceles as well. X 20 T 30˚ A

  6. Corollary 9-13.3 Statement Reason Un-named. Y 1. RSU  ( R + RSQ). 1. Given < < B The exterior angle of any triangle, has the same degrees as the two remote interior angles added together. < A C E 2. Corollary 9-13.3 2. TQR  ( R + RSQ). < < X Z < 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. 3. TQR  RSU. 3. TPE < < 4. TQR is supp. to SQR. 4. Supp. Pos. < < 5. USR is supp. to QSR. 5. Supp. Pos. < < Triangles Problem #2 6. Supp. Theorem. 6. QSR  SQR. < < 7. Def of Isosc.  7. RSQ is isosc. Use figure to below. Given: RSU  ( R + RSQ). < < < R Prove: RSQ is isosceles. T U Q S

  7. Quadrilateral Square Definitions Rectangle Quadrilateral A parallelogram with 4 right angles and 4 congruent sides. Any parallelogram with 4 right angles. Any 2 dimensional figure with exactly 4 sides. Rhombus Parallelogram Trapezoid Any quadrilateral with only one pair of opposite sides being parallel. Any parallelogram with 4 congruent sides. Any quadrilateral with every pair of opposite sides being parallel.

  8. Theorem 9-16 Un-named B C In a parallelogram, the opposite angles are congruent. Restatement Given: Parallelogram ABCD, then A  C & B  D. A < < < < D B F C Quadrilaterals Problem #1 Use the figure to the right. Given: Parallelogram ABCD with FB = HD & BE = GD. G Statement Reason E 1. ABCD is a parallelogram; BE = GD; FB = HD 1. Given A H D Prove: EF = GH. 2. B  D. 2. 9-16 < < 3. FBE  HDG 3. SAS 4. EF = GH 4. CPCTC

  9. Theorem 9-25 B Un-named If a quadrilaterals diagonals are perpendicular to each other and bisect each other, then the quadrilateral is a rhombus. A C Restatement Given: ABCD, with AC  BD, and AC and BD bisecting each other, then it is a rhombus. D B A Quadrilateral Problem #2 Use the figure on the right. Given: ABCD , with AC  BD, and AC and BD bisecting each other. C Statement Reason D 1. Given 1. ABCD , with AC  BD, and AC and BD bisecting each other. Prove: All sides are equal. 2. ABCD is a rhombus. 2. 9-25 3. All sides are equal. 3. Def of rhombus

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