Chapter 9 Summary Project

By Matthew Donoghue

Starring:Parallelism

Triangles

&

Quadrilaterals

Period 2

12/17

P a r a e i s mDefinitions

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.

4. ACB CBD

Theorem 9-5The 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

=

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.

RestatementGiven: L1 L2 L3. All three lines are intersected by transversals T1 and T2; AB = BC. Then EF = FG.

= =

Parallelism Problem #2Use 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.

TrangleDefinitions 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 30-60-90Triangle TheoremIn 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.

RestatementGiven: ABC is a right triangle; A has a measure of 30˚; D is the midpoint of AB. Then BC = 1/2 AB.

Triangles Problem #1Use the. figure on the left.Given: MAT and ROX are right triangles; MA = RX; R & T = 30˚; MT = 20.< <

Statement Reason

1. MAT and ROXare right triangles;MT = 20; R & T = 30˚

1. Given

< <

2. MA = 10 2. 30-60-90

3. Given3. MA = RX

4. RX = 10 4. Substitution

5. 30-60-905. OX = 5

Corollary 9-13.3Un-named.

The exterior angle of any triangle, has the same degrees as the two remote interior angles added together.

RestatementGiven: 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 #2Use figure to below.Given: RSU ( R + RSQ).<<

Prove: RSQ is isosceles.

<

1. RSU ( R + RSQ).

<<

1. Given

Statement Reason

2. TQR ( R + RSQ).

<<

2. Corollary 9-13.3

3. TQR RSU.

<

<

< < 3. TPE4. 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.

QuadrilateralDefinitions

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.

Theorem 9-16Un-namedIn a parallelogram, the opposite angles are congruent.

RestatementGiven: 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

Statement Reason1. Given

2. 9-162. B D.

3. SAS

4. CPCTC

< <3. FBE HDG

4. EF = GH

Theorem 9-25Un-named

If a quadrilaterals diagonals are perpendicular to each other and bisect each other, then the quadrilateral is a rhombus.

RestatementGiven: ABCD, with AC � BD, and AC and BD bisecting each other, then it is a rhombus.

Quadrilateral Problem #2Use the figure on the right.Given: ABCD , with AC � BD, and AC and BD bisecting each other. Statement Reason

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.