1. By: Tyler Register and Tre Burse
without 4.7
Proving quadrilateral properties
Conditions for special quadrilaterals
Constructing transformations
By: Tyler Register
and
Tre Burse
geometry

2. The Vocabulary and Theorems
A diagonal of a parallelogram divides the parallelogram into two equal triangles
Opposite sides of a parallelogram are congruent
Opposite angles of a parallelogram are congruent
Diagonals of a parallelogram bisect each other

3. Theorems cont. A rhombus is a parallelogram
A rectangle is a parallelogram
The diagonals and sides of a rhombus form 4 congruent triangles
The diagonals of a rhombus are perpendicular
The diagonals of a rectangle are congruent
A square is a rhombus

4. Theorems cont.
The diagonals of a square are perpendicular and are bisectors of the angles
If two pairs of opposite sides of a quadrilateral are congruent then the quadrilateral is a parallelogram
If the diagonals of a quadrilateral bisect each other then the quadrilateral is a parallelogram

5. Theorems
If one angle of a parallelogram is a right angle then the parallelogram is a rectangle
If the diagonals of a parallelogram are congruent then the parallelogram is a rectangle
If one pair of adjacent sides of a quadrilateral are congruent then the quadrilateral is a rhombus

6. More Theorems
If the diagonals of a parallelogram bisect the angles of the parallelogram then it is a rhombus
If the diagonals of a parallelogram are perpendicular than it is a rhombus
Triangle mid-segment theorem- A mid-segment of a triangle is parallel to a side of the triangle and its length is equal to half the length of than side

7. The Last Theorem Slide
Betweenness postulate- given the three points: P, Q, and R PQ+QR=PR then Q is between P and R on a line.
The Triangle inequality theorem- The sum of any two sides of a triangle are greater than the other side.

8. 4-5
Statements Reasons
PLGM is a parallelogram and LM is a diagonal
Given
Def of parallelogram
Objective- Prove quadrilateral conjectures by using triangle congruence postulates and theorems.
PL II GM
Given: parallelogram PLGM with diagonal LM
< 3 = < 2
Alt. Int. angles
PM II GL
Def of parallelogram
Prove: triangle LGM= triangle MPL
<1 = <4
Alt. Int. angles
LM=LM
reflexive
LGM= MPL
ASA
P L
M G

9. Conditions of special quadrilaterals
4-6
Conditions of special quadrilaterals
There are many theorems in this section that state special cases in quadrilaterals
The most notable of these theorems is the House Builder Theorem
There is also the Triangle Mid- segment Theorem
House Builder Theorem: If the diagonals of a parallelogram are congruent then the parallelogram is a rectangle
Triangle Mid-segment Theorem: A mid-segment of a triangle is parallel to a side of the triangle and its length is equal to half the length of than side
The list of the theorems in 4-6 are on page 5 and 6

10. 4-8 Constructing transformations
This section has one theorem and one postulate
The Betweenness postulate (converse of the segment addition postulate) and the Triangle Inequality Theorem
The Betweenness postulate: given the three points: P, Q, and R PQ+QR=PR then Q is between P and R on a line.
Triangle Inequality Theorem: The sum of any two sides of a triangle are greater than the other side.
5+7>X
X+5>7
X+7>5
2<X<13 5 7 X

11. Quiz Which of the following are possible lengths of a triangle?
A. 14,8, B.16,7, C.18,8,24
If one angle of a quadrilateral is a right angle than the quadrilateral is a ___________.
Find the measure of the following angles:
<Q=
<RPQ=
<PRQ=
Rectangle 60
P Q
S R 40 80 60 40