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

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

3. 4. ACB  CBD 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 = << Use Diagram above right. Given: AC = BD, AB = CD. Lines L1 and L2 are cut by a transversal T. Prove L1 L2. Parallelism Problem #1. = << Statement Reason 1. AC = BD, AB = CD 2. CB = CB 3. ∆ABC  ∆DCB 5.  L1 L2 1. Given 3. SSS 4. CPCTC 5. AIP 2. RPE =

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

5. TrangleTrangle Definitions Theorems Isosceles Triangle 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. Right Triangle A right triangle has one right and two acute angles. P.S. Right triangles can be isosceles as well. < 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 Given:  ABC is a right triangle; A has a measure of 30˚; D is the midpoint of AB. Then BC = 1/2 AB. Triangles Problem #1 Use the. figure on the left. Given:  MAT and  ROX are right triangles; MA = RX; R & T = 30˚; MT = 20. << Statement Reason 1.  MAT and  ROX are right triangles; MT = 20; R & T = 30˚ 1. Given << 2. MA = 10 2. 30-60-90 3. Given3. MA = RX 4. RX = 104. Substitution 5. 30-60-905. OX = 5

6. 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. < 1. RSU  ( R + RSQ). < < 1. Given StatementReason 2. TQR  ( R + RSQ). < < 2. Corollary 9- 13.3 3. TQR  RSU. < < << 3. TPE 4. TQR is supp. to SQR. < < 5. USR is supp. to QSR. < < 4. Supp. Pos. 5. Supp. Pos. 6. QSR  SQR. << 6. Supp. Theorem. 7.  RSQ is isosc. 7. Def of Isosc. 

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

8. 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. Prove: EF = GH. 1. ABCD is a parallelogram; BE = GD; FB = HD StatementReason 1. Given 2. 9-16 2. B  D. 3. SAS 4. CPCTC << 3.  FBE   HDG 4. EF = GH

9. 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. StatementReason 1. ABCD, with AC  BD, and AC and BD bisecting each other. 1. Given 2. 9-252. ABCD is a rhombus. 3. All sides are equal.3. Def of rhombus Prove: All sides are equal.