Special Quadrilaterals Constructions and Proofs
Special Quadrilaterals
Constructions and Proofs
Proofs: Pg. 272: 9
Show: BN bisects B BN bisects N
Given: Kite BENY with BE BY EN YN
43
21
E
B
Y
N
Construct Kite BENY
1. Draw BE 2. Make B3. Bisect B4. Copy BE on the other side. Name the endpoint Y5. Pick a point on the angle bisector. Name the point N.6. Construct EN and YN
Y
BE
N
Flow chart proof
BN bisects B and BN bisects N Definition of angle bisector
1 2 and3 4CPCTC
BEN BYNSSS SSS
BN BNSame Seg.
EN YNGiven
BE BYGiven
Paragraph Proof
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 =.
Pg. 272: 10
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.
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.
Construct a kite and name it.
1. Draw BE2. Make B3. Bisect B4. Copy BE on the other side of B. Name the endpoint Y.5. Pick a point on the bisector and name it N.6. Construct EN and YN7. Construct EY
D
Y
B
E
N
Flowchart Proof
BN is the bisector of EY
Def. of bisector
EY BN adjacent s
BN bisects EYD is the midpt
EDB YDBED DY
CPCTC
Quad. BENYis a Kite.Given
BED BYDSAS SAS
BD BDSame Seg.
EBD YBDDiagonal bisectsvertex angles
BE BYDef. of a Kite
Paragraph Proof
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. ΔBED = ΔBYD because SAS = SAS. EDB = YDB and ED = EY because corresponding parts to = Δs 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.
Pg. 278: 8
Show: LN RD
Given: Midsegment LN in FOA Midsegment RD in IOA
R D
LN
I
F
AO
Construct given
1. Make FOA with O as an obtuse .2. Reflect FOA over OA. Name the new point I.3. Construct the midpoints of FO, FA, IO, and IA. Name the midpoints L, N, R, D respectively.4. Construct the midsegments. DR
NL
I
F
AO
Flowchart Proof
LN RD2 seg. to the same segare to each other.
RD OAMIdseg. of a
LN OAMIdseg. of a
RD is a midsegment of IOAGiven
LN is a midsegment of FOAGiven
Paragraph Proof
LN is the midsegment ofΔFOA and RD is the midsegment of ΔIOA 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.
Pg. 284: 13
Given: LEAN is a parallelogram
Show: EN and LA bisect each other
T
N
E
A
L
Construct the given
1. Draw EA2. Pick a point not on EA and construct a line to EA. 3. Copy EA onto the parallel line and name it LN.4. Construct LE and AN.5. Construct LA and EN, naming their intersection T.
T
EA
LN
Flowchart Proof
EN and LA bisecteach otherDef. seg. bisectorAT LT
CPCTC
ET NTCPCTC
AET LNTASA
EAL NLAAIA
AEN LNEAIA
AE LNOpp. sides
EA LN Def. of parallelogramLEAN is a
parallelogramGiven
Paragraph proof
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 ΔAET = Δ LNT because ASA = ASA. ET = NT and AT = LT because corresponding parts of = Δs are =. Therefore, EN and LA bisect each other because each divides the other into 2 = parts.
Pg. 296: 26
4
32
1Show: QUAD is a rhombus
Given: Quadrilateral QUAD with QU UA AD DQ
Q
U
DA
Construct the given
1. Draw DA2. Make D3. Copy DA on the other side of D. Name the endpoint Q.4. Construct a line to DA5. Copy DA onto the line, with one endpoint at Q and the other endpoint named U6. Construct UA.7. Construct diagonal DU.
Q
D
A
U
Flowchart proof
QUAD is a rhombusDef. of a rhombus
QU UA AD DQGiven
QUAD is a paralleogramDef. of parallelogram
QU ADQD AUCon of lines
QUD ADUQDU DUACPCTC
QUD ADUSSS
DU DUsame seg
QD AUGiven
QU ADGiven
Paragraph proof
QU = AD and QD = AU because they are given =. DU = DU because they are the sane segment. ΔQUD = ΔADU because SSS = SSS. QUD = ADU and QDU = DUA because they are corresponding parts = Δs 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.
Pg. 307: 27
Given: Rhombus DENI with diagonal DN
Show: Diagonal DN bisects D and N
4
3
21
E
I N
D
Pg. 307: 28
Construct a diagram to prove the Parallelogram Opposite Sides Conjecture.
Homework: Due Thursday
Pg. 300 – 302: 1 – 9
Make sure you construct the given before you write the proof.