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# Special quadrilaterals proofs ans constructions

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### Special quadrilaterals proofs ans constructions

1. 1. Special Quadrilaterals Constructions and Proofs
2. 2. Proofs: Pg. 272: 9
3. 3. Construct Kite BENY
4. 4. Flow chart proof
5. 5. Paragraph Proof <ul><li>BE = BY and EN = YN because they are given true. BN = BN because they are the same segment. ΔBEN = ΔBYN because if the three sides of Δ are = to three sides of another Δ, then the Δs are =.  1 =  2 and  3 =  4 because corresponding parts of = Δs are =. </li></ul>
6. 6. Pg. 272: 10 <ul><li>Write a paragraph proof of the Kite Diagonal Bisector Conjecture. Either show how it follows logically from the Kite Bisector conjecture or from the converse of the Perpendicular Bisector Conjecture. </li></ul>
7. 7. Kite Diagonal Conjecture : If a quadrilateral is a kite, then the diagonal connecting the vertex angles of the kite is the perpendicular bisector of the other diagonal. <ul><li>Construct a kite and name it. </li></ul>
8. 8. Flowchart Proof
9. 9. Paragraph Proof <ul><li>Quadrilateral BENY is a Kite because it is given one. BE = EY because the consecutive sides of a kite are =.  EBD =  YBD because the diagonal that connects the vertex  s bisects the vertex  s. BD is = to itself because any segment is = to itself. Δ BED = Δ BYD because SAS = SAS.  EDB =  YDB and ED = EY because corresponding parts to = Δ s are =. BN  EY because the two adjacent supplementary angles are =. D is the midpoint of EY because it divides it into two = segments. That makes BN the  bisector of EY. </li></ul>
10. 10. Pg. 278: 8
11. 11. Construct given
12. 12. Flowchart Proof
13. 13. Paragraph Proof <ul><li>LN is the midsegment of Δ FOA and RD is the midsegment of Δ IOA because it is given true. LN is ↑↑ to OA and RD is ↑↑ to OA because a midsegment of a Δ is ↑↑ to the third side of the Δ . LN is ↑↑ to RD because two segments ↑↑ to the same segment are ↑↑ to each other. </li></ul>
14. 14. Pg. 284: 13
15. 15. Construct the given
16. 16. Flowchart Proof
17. 17. Paragraph proof <ul><li>Quad LEAN is a parallelogram because it is given so. Because LEAN is a parallelogram, EA ↑↑ LN. AE = LN because opposite sides of a parallelogram are =.  AEN =  LNE and  EAL =  NLA because alternate interior angles of parallel lines are =. That makes Δ AET = Δ LNT because ASA = ASA. ET = NT and AT = LT because corresponding parts of = Δ s are =. Therefore, EN and LA bisect each other because each divides the other into 2 = parts. </li></ul>
18. 18. Pg. 296: 26
19. 19. Construct the given
20. 20. Flowchart proof
21. 21. Paragraph proof <ul><li>QU = AD and QD = AU because they are given =. DU = DU because they are the sane segment. Δ QUD = Δ ADU because SSS = SSS.  QUD =  ADU and  QDU =  DUA because they are corresponding parts = Δ s are =. QU = AD and QD = AU because alternate interior  s = make ↑↑ . QUAD is a parallelogram because a quadrilateral with opposite sides ↑↑ is a parallelogram. QU = AD = QD = AU because they are given =. QUAD is a rhombus because a parallelogram that is equilateral is a rhombus. </li></ul>
22. 22. Pg. 307: 27
23. 23. Pg. 307: 28 <ul><li>Construct a diagram to prove the Parallelogram Opposite Sides Conjecture. </li></ul>
24. 24. Homework: Due Thursday <ul><li>Pg. 300 – 302: 1 – 9 </li></ul><ul><li>Make sure you construct the given before you write the proof. </li></ul>