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.