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**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

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**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

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**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

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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

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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

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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

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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.

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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

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**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

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**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

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**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

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6.4 Rhombuses, Rectangles, and Squares Day 4 Review Find the value of the variables. 52° 68° h p (2p-14)° 50° 52° + 68° + h = 180° 120° + h = 180 °

6.4 Rhombuses, Rectangles, and Squares Day 4 Review Find the value of the variables. 52° 68° h p (2p-14)° 50° 52° + 68° + h = 180° 120° + h = 180 °

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