Download Use figure below right. Given

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
Trangle
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