Given: ABC CD bisects AB CD AB Prove: ACD BCDmwebster23.weebly.com/uploads/5/5/2/9/55298051/...#1...

Given: ABC CD bisects AB

Prove: ACD BCD

Statement

1. ABC CD bisects AB

2. AD DB Side

3. CDA and CDB are right

4. CDA CDB Angle

5. CD CD Side

6. ACD BCD

Reasons 1. Given 2. A bisector cuts a segment into 2

parts.

3. lines form right .

4. All rt are . 5. Reflexive post.

6. SAS SAS

Given: ABC and DBE bisect each other.

Prove: ABD CBD

Statement 1. ABC and DBE bisect each other.

2. AB BC Side

BD BE Side

3. ABD and BEC are vertical

4. ABD BEC Angle

5. ABD CBD

Reasons 1. Given 2. A bisector cuts a segment into 2

parts.

3. Intersecting lines form vertical .

4. Vertical are .

5. SAS SAS

Given: AB CD and BC DA

DAB, ABC, BCD and CDA

are rt

Prove: ABC ADC

Statement

1. AB CD Side

BC DA Side

2. DAB, ABC, BCD and CDA

are rt

3. ABC ADC Angle

4. ABC ADC

Reasons 1. Given 2. Given

3. All rt are .

4. SAS SAS

Given: PQR RQS

Prove: PQR RQS

Statement

1. PQR RQS Angle

PQ QS Side

2. RQ RQ Side

3. PQR RQS

Reasons 1. Given 2. Reflexive Post.

3. SAS SAS

Given: AEB & CED intersect at E E is the midpoint AEB

AC AE & BD BE

Prove: AEC BED

Statement 1. AEB & CED intersect at E E is the midpoint AEB

AC AE & BD BE 2. AEC and BED are vertical

3. AEC BED Angle

4. AE EB Side

5. A & B are rt.

6. A B Angle

7. AEC BED

Reasons 1. Given

2. Intersecting lines form vertical .

3. Vertical are . 4. A midpoint cut a segment into 2

5. lines form right .

6. All rt are .

7. ASA ASA

Given: AEB bisects CED

AC CED & BD CED

Prove: EAC EBD

Statement 1. AEB bisects CED

AC CED & BD CED

2. CE ED Side

3. ACE & EDB are rt

4. ACE EDB Angle

Reasons 1. Given 2. A bisector cuts an angle into

2 parts.

3. Lines form rt .

4. All rt are

5. AEC & DEB are vertical

6. AEC DEB Angle

7. EAC EBD

5. Intersect lines form vertical

6. Vertical are

7. ASA ASA

Given: ABC is equilateral D midpoint of AB

Prove: ACD BCD

Statement

1. ABC is equilateral D midpoint of AB

2. AC BC Side

3. AD DB Side

4. CD CD Side

5. ACD BCD

Reasons 1. Given

2. All sides of an equilateral are 3. A midpoint cuts a segment into

2 parts. 4. Reflexive Post

5. SSS SSS

Given: mA = 50, mB = 45,

AB = 10cm, mD = 50

mE = 45 and DE = 10cm

Prove: ABC DEF

Statement

1. mA = 50, mB = 45,

AB = 10cm, mD = 50

mE = 45 and DE = 10cm

2. A = D Angle and

B = E Angle AB = DE Side

3. ABC DEF

Reasons 1. Given 2. Transitive Prop

3. ASA ASA

Given: GEH bisects DEF

mD = mF

Prove: GFE DEH

Statement 1. GEH bisects DEF

mD = mF Angle

2. DE EF Side

3. 1 & 2 are vertical

4. 1 2 Angle

5. GFE DEH

Reasons 1. Given

2. Bisector cut a segment into 2 parts.

4. Vertical are

5. ASA ASA

Given: PQ bisects RS at M

Prove: RMQ SMP

Statement 1. PQ bisects RS at M

R S Angle

2. RM MS Side

Reasons 1. Given

2. Bisector cut a segment into 2

3. 1 & 2 are vertical angles

4. 1 2 Angle

5. RMQ SMP

4. Vertical are

5. ASA ASA

Given: DE DG

Prove: DEF DFG

Statement

1. DE DG Side

EF GF Side

2. DF DF Side

3. DEF DFG

Reasons 1. Given 2. Reflexive Post

3. SSS SSS

Given: KM bisects LKJ

Prove: JKM LKM

Statement 1. KM bisects LKJ

LK JK Side

2. 1 2 Angle

Reasons 1. Given

2. An bisectors cuts the into

2 parts

3. KM KM Side

4. JKM LKM

3. Reflexive Post

4. SAS SAS

Given: . PR QR

P Q RS is a median

Prove: PSR QSR

Statement

1. PR QR Side

P Q Angle RS is a median Side

2. PS SQ

3. PSR QSR

Reasons 1. Given 2. A median cuts the side into

2 parts

3. SAS SAS

Given: EG is bisector EG is an altitude

Prove: DEG GEF

Statement

1. EG is bisector EG is an altitude

2. 3 4 Angle

Reasons 1. Given

2. An bisector cuts an into

2 parts.

3. EG DF

4. 1 & 2 are rt

5. 1 2 Angle

6. GE GE Side

7. DEG GEF

3. An altitude form lines.

4. lines form right angles.

5. All right angles are 6. Reflexive Post

7. ASA ASA

Given: A and D are a rt

Prove: EC FB

Statement

1. A and D are a rt

AE DF Side

2. A D Angle

3. BC BC

4. AB + BC CD + BC

or AC BD Side

5. AEC DFB

6. EC FB

Reasons 1. Given

2. All right angles are . 3. Reflexive Post. 4. Addition Prop.

5. SAS SAS

6. Corresponding parts of are .

Given: CA CB D midpoint of AB

Prove: A B

Statement

1. CA CB Side D midpoint of AB

Reasons 1. Given

2. AD DB Side

3. CD CD Side

4. ADC DBC

5. A B

2. A midpoint cuts a segment into

2 parts 3. Reflexive Post 4. SSS SSS

Given: . AB CD

CAB ACD

Prove: AD CB

Statement

1. AB CD Side

CAB ACD Angle

2. AC AC Side

3. ACD ABC

4. AD CB

Reasons 1. Given 2. Reflexive Post 3. SAS SAS

Given: AEB & CED bisect each

Prove: C D

Statement 1. AEB & CED bisect each other

2. CE ED Side & AE EB Side

3. 1 and 2 are vertical

Reasons 1. Given 2. A bisector cuts segments into

2 parts.

4. 1 2 Angle

5. AEC DEB

6. C D

4. Vertical are

5. SAS SAS

6. Corresponding parts of are

Given: KLM & NML are rt

Prove: K N

Statement

1. KLM & NML are rt

KL NM Side

2. KLM NML Angle

3. LM LM Side

4. KLM LNM

5. K N

Reasons 1. Given

2. All rt are 3. Reflexive Post.

4. SAS SAS

Given: AB BC CD

PA PD & PB PC

Prove: a) APB DPC

b) APC DPB

Statement

1. AB BC CD Side

PA PD Side & PB PC Side

2. ABP CDP

3. APB DPC

Reasons 1. Given

2. SSS SSS

4. BPC BPC

5. APB + BPC DPC + BPC

or APC DPB

4. Reflexive Post. 5. Addition Prop.

Given: PM is Altitude PM is median

Prove: a) LNP is isosceles

b) PM is bisector

Statement 1. PM is Altitude & PM is median

2. PM LN

3. 1 and 2 are rt

4. 1 2

5. LM MN

6. PM PM

7. LMP PMN

8. PL PN

9. LNP is isosceles

10. LPN MPN

11. PM is bisector

Reasons 1. Given

2. An altitude form lines.

3. lines form right angles.

4. All right angles are 5. A median cuts the side into

2 parts 6. Reflexive Post.

7. SAS SAS

9. An Isosceles is a with2 sides

10.Corresponding parts of are .

11. A bisector cuts an into

2 parts

Given: CA CB

Prove: CAD CBE

Statement

1. CA CB

2. 2 3

3. 1 & 2 are supplementary

3 & 4 are supplementary

4. 1 4 or CAD CBE

Reasons 1. Given

2. If 2 sides are then the opposite

3. Supplementary are form by a linear pair.

4. Supplement of are .

Given: AB CB & AD CD

Prove: BAD BCD

Statement

1. AB CB & AD CD

2. 1 2

3. 1 + 3 2 + 4

or BAD BCD

Reasons 1. Given

2. If 2 sides are then the opposite

are . 3. Addition Post.

Given: ΔABC ΔDEF M is midpoint of AB N is midpoint DE

Prove: ΔAMC ΔDNF

Statement

1. ΔABC ΔDEF 2. M is midpoint of AB N is midpoint DE

3. D A Angle and DF AC Side

4. AM MB and DN NE Side

5. ΔAMC ΔDNF

Reasons 1. Given 2. Given

3. Corresponding parts of Δ are 4. A midpoint cuts a segment into

2 parts

5. SAS SAS

Given: ΔABC ΔDEF

CG bisects ACB

FH bisects DFE

Prove: CG FH

Statement

1. ΔABC ΔDEF

CG bisects ACB

FH bisects DFE

Reasons

Given: ΔAME ΔBMF

Prove: AD BC

Statement

1. ΔAME ΔBMF

2. EM MF

AM MB Side

1 2 Angle

3. DE + EM CF + MF

or DM MC Side

4. ΔADM ΔBCM

5. AD BC

Reasons 1. Given

2. Corresponding parts of Δ are 3. Addition Post.

4. SAS SAS

5. Corresponding parts of Δ are

Given: AEC & DEB bisect each other Prove: E is midpoint of FEG

Statement 1. AEC & DEB bisect each other

Reasons 1. Given

2. DE BE Side and AE EC Side

3. AEB & DEC are vertical

4. AEB DEC Angle

5. ΔAEB ΔDEC

6. D B

7. 1 & 2 are vertical angles

8. 1 2

9. ΔGEB ΔDEF

10. GE FE 11. E is midpoint of FEG

2. A bisector cuts a segment into

2 parts.

3. Intersecting lines form vertical

4. Vertical are .

5. SAS SAS

7. Intersecting lines form vertical

8. Vertical are .

9. ASA ASA

10. Corresponding parts of Δ are 11. A midpoint divides a segment

into 2 parts.

Given: BC BA

BD bisects CBA

Prove: DB bisects CDA

Statement Reasons

1. BC BA Side BD bisects CBA

2. 1 2 Angle

3. BD BD Side

4. ΔABD ΔBCD

5. 3 4

6. DB bisects CDA

1. Given 2. A bisector cuts an angle into

2 parts. 3. Reflexive Post.

4. SAS SAS

5. Corresponding parts of Δ are 6. A angle bisector cuts an angle

into 2 parts.

Given: AE FB

A and B are Rt.

Prove: ADF CBE

Statement

1. AE FB

DA CB Side

A and B are Rt.

2. EF EF

3. AE + EF FB + EF

or AF EB Side

Reasons 1. Given 2. Reflexive Post 3. Addition Property

4. A B Angle

5. ADF CBE

6. DF CE

4. All rt. are .

5. SAS SAS

Given: SPR SQT

Prove: SRQ STP

Statement

1. SPR SQT Side

2. S S Angle

3. SPR – PR SQT – QT

or SR ST Side

4. SRQ STP

5. R T

Reasons 1. Given 2. Reflexive Post 3. Subtraction Property

4. SAS SAS

Given: DA CB

DA AB & CB AB

Prove: DAB CBA

Statement

1. DA CB Side

DA AB & CB AB

2. DAB and CBA are rt

3. DAB CBA Angle

4. AB AB Side

5. DAB CBA

6. AC BD

Reasons 1. Given

2. lines form rt .

3. All rt are . 4. Reflexive post.

5. SAS SAS

6. Corresponding parts of Δ are .

Given: BAE CBF

BCE CDF

Prove: AE BF

Statement

1. BAE CBF Angle

BCE CDF Angle

2. BC BC

3. AB + BC CD + BC

or AC BD Side

4. AEC BDF

5. AE BF

Reasons 1. Given 2. Reflexive Post. 3. Addition Property.

4. ASA ASA

Given: TM TN M is midpoint TR N is midpoint TS

Prove: RN SM

Statement Reasons

1. TM TN Side M is midpoint TR N is midpoint TS

2. T T Angle 3. RM is ½ of TR NS is ½ of TS

4. RM NS

5. TM + RM TN + NS

or RT TS Side

6. RTN MTS

7. RN SM

1. Given 2. Reflexive Post. 3. A midpoint cuts a segment in .

4. ½ of parts are . 5. Addition Property

6. SAS SAS

Given: AD CE & DB EB

Prove: ADC CEA

Statement

1. AD CE & DB EB Side

Reasons 1. Given

2. B B Angle

3. AD + DB CE + EB

or AB BC Side

4. ABE BCD

5. 1 2

6. 1 & 3 are supplementary

2 & 4 are supplementary

7. 3 4 or

ADC CEA

2. Reflexive Post 3. Addition Post.

4. SAS SAS

6. A st. line forms supplementary .

7. Supplements of are .

Given: AE BF & AB CD

ABF is the suppl. of A

Prove: AEC BFD

Statement

1. AE BF Side & AB CD

ABF is the suppl. of A

Reasons 1. Given

2. A 1 Angle

3. BC BC

4. AB + BC CD + BC

or AC BD Side

5. AEC BFD

2. Supplements of are . 3. Reflexive Post. 4. Addition Property.

5. SAS SAS

Given: AB CB

BD bisects ABC

Prove: AE CE

Statement

1. AB CB Side

BD bisects ABC

2. 1 2 Angle

3. BE BE Side

4. BEC BEA

5. AE CE

Reasons 1. Given

2. A bisector cuts an into

2 parts. 3. Reflexive Post.

4. SAS SAS

Given: PB PC

Prove: ABP DCP

Statement

1. PB PC

Reasons 1. Given

2. 1 2

3. 1 & ABP are supplementary

2 & DCP are supplementary

4. ABP DCP

2. opposite sides are .

3. Supplementay are formed by a linear pair.

Given: AC and BD are bisectors of each other.

Prove: AB BC CD DA

Statement

1. AC and BD are bisectors of each other

2. 1, 2, 3 and 4 are rt

3. 1 2 3 4 Angle

4. AE EC and BE DE 2 sides

5. ABE BEC DEC AED

6. AB BC CD DA

Reasons 1. Given

2. lines form rt .

3. All rt are . 4. A bisector cuts a segment into

2 parts.

5. SAS SAS

Given: AEFB, 1 2

CE DF, AE BF

Prove: AFD BEC

Statement Reasons

1. AEFB, 1 2 Angle

CE DF Side, AE BF

2. EF EF

3. AE + EF BF + EF or

AF EB Side

4. AFD BEC

1. Given 2. Reflexive Post. 3. Addition Property

4. SAS SAS

Given: SX SY, XR YT

Prove: RSY TSX

Statement

1. SX SY Side, XR YT

2. SX + XR SY + YT

or SR ST Side

3. S S Angle

4. RSY TSX

Reasons 1. Given 2. Addition Post. 3. Reflexive Post.

4. SAS SAS

Given: DA CB

DA AB, CB AB

Prove: DAB CBA

Statement

1. DA CB Side

DA AB, CB AB

2. DAB and CBA are rt.

3. DAB CBA Angle

4. AB AB Side

5. DAB CBA

Reasons 1. Given

2. lines form rt

3. All rt. are 4. Reflexive Post.

5. SAS SAS

Given: AF EC

1 2, 3 4

Prove: ABE CDF

Statement

1. AF EC

1 2, 3 4 Angle

2. DFC BEA Angle

3. EF EF

4. AF + EF EC + EF or

AE FC Side

5. ABE CDF

Reasons 1. Given

2. Supplements of are 3. Reflexive post. 4. Addition Post.

5. AAS AAS

Given: AB BF, CD BF

1 2, BD FE

Prove: ABE CDF

Statement

1. AB BF, CD BF

1 2 Side , BD FE

2. B and CDF are rt.

3. B CDF Angle

4. DE DE

5. BD + DE FE + DE or

BE DF Side

6. ABE CDF

Reasons 1. Given

2. lines form rt.

3. All rt. are 4. Reflexive Post. 5. Addition Post.

6, ASA ASA

Given: BAC BCA

CD bisects BCA

AE bisects BAC

Prove: ADC CEA

Statement

1. BAC BCA Angle

CD bisects BCA

AE bisects BAC

2. ECA ½BAC and

DCA ½BCA

3. ECA DCA Angle

4. AC AC Side

5. ADC CEA

Reasons 1. Given

2. bisector cuts an in ½

3. ½ of are 4. Reflexive post.

5. ASA ASA

#1 #45

Given: TR TS, MR NS

Prove: RTN STM

Statement

1. TR TS Side, MR NS

2, TR – MR TS – NS or

TM TN Side

3. T T Angle

4. RTN STM

Reasons 1. Given 2. Subtraction Post. 3. Reflexive Post.

4. ASA ASA

Given: CEA CDB, ABC AD and BE intersect at P

PAB PBA

Prove: PE PD

Statement

1. CEA CDB, ABC AD and BE intersect at P

PAB PBA 2.

Reasons 1. Given

Given: AB AD and BC DC

Prove: 1 2

Statement

1. AB AD and BC DC

2. AC AC

3. ABC ADC

4. AE AE

5. BAE DAE

6. ABE ADE

7. 1 2

Reasons 1. Given 2. Reflexive Post.

3. SSS SSS 4. Reflexive Post.

6. SAS SAS

Given: BD is both median and

altitude to AC

Prove: BA BC

Statement 1. BD is both median and

altitude to AC

2. AD CD Side

3. ADB and CDB are rt.

4. ADB CDB Angle

5. BD BD Side

6. ABD CBD

Reasons 1. Given

2. A median cuts a segment into 2 parts

3. Lines form rt.

7. BA BC

6. SAS SAS

Given: CDE CED and AD EB

Prove: ACC BCE

Statement

1. CDE CED and AD EB Side

2. CDA CEB Angle

3. CD CE Side

4. ADC BEC

5. ACD BCE

Reasons 1. Given

3. Sides opp. in a are

4. SAS SAS

Given: Isosceles triangle CAT

CT AT and ST bisects CTA

Prove: SCA SAC

Statement 1. Isosceles triangle CAT

CT AT Side and ST bisects CTA

2. CTS ATS Angle

3. ST ST Side

4. CST AST

Reasons 1. Given

2. An bisector cuts an into 2 parts 3. Reflexive Post.

4. SAS SAS

5. CS AS

6. SCA SAC

6. opp. sides in a are

Given: 1 2

Prove: ABD CBD

Statement

1. 1 2 and DB AC

2. DBA and DBC are rt.

3. DBA DBC Angle

4. DAB DCA Angle

5. DB DB Side

6. ABD CBD

Reasons 1. Given

2. lines form rt.

3. All rt. are

4. Supplements of are 5. Reflexive Post.

6. AAS AAS

Given: P S R is midpoint of PS

Given: PQR STR

Statement

1. P S Angle R is midpoint of PS

2. PR RS Side

3. QRP and TRS are vertical

Reasons 1. Given

2. A midpoint cuts a segment into 2 parts

3. Intersecting lines form vert.

4. QRP TRS Angle

5. PQR STR

4. Vertical are

5. ASA ASA

Given: FG DE G is midpoint of DE

Given: DFG EFG

Statement

1. FG DE G is midpoint of DE

2. FGD and FGE are rt.

3. FGD FGE Angle

4. FG FG Side

5. DG GE Side

6. DFG EFG

Reasons 1. Given

2. lines form rt.

5. A midpoint cuts a segment into 2 parts.

6. SAS SAS

Given: AC CB D is midpoint of AB

Prove: ACD BCD

Statement

1. AC CB Side D is midpoint of AB

Reasons 1. Given

2. AD DB Side

3. CD CD Side

4. ACD BCD

2. A midpoint cuts a segment into 2 parts. 3. Reflexive Post.

4. SSS SSS

Given: PT bisects QS

PQ QS and TS QS

Prove: PQR RST

Statement 1. PT bisects QS

PQ QS and TS QS

2. QR RS Side

3. PRQ and TRS are vertical

4. PRQ TRS Angle

5. Q and S are rt.

6. Q S Angle

7. PQR RST

Reasons 1. Given

2. A bisector cuts a segment into 2 parts.

4. All vert. are

5. lines form rt.

6. All rt. are

7. ASA ASA

Given: AB ED and FE CB

FE AD and CB AD

Prove: AEF CBD

Statement

1. AB ED and FE CB Side

Reasons 1. Given

FE AD and CB AD

2. BE BE

3. AB + BE ED + BE or

AE DB Side

4. AEF and DBF are rt.

5. AEF DBF Angle

6. AEF CBD

2. Reflexive Post. 3. Addition Post.

4. lines form rt.

5. All rt. are

6. SAS SAS

Given: SM is bisector of LP

Prove: RLM QPM

Statement

1. SM is bisector of LP

RM MQ Side

2. SML and SMP are rt.

3. 1 2 Angle

4. LM PM Side

5. RLM QPM

Reasons 1. Given

2. lines form rt.

3. Complements of are

5. SAS SAS

Given: AC BC

Prove: ACD BCD

Statement Reasons

1. AC BC

2. CDA and CDB are rt.

3. CDA CDB

4. CD CD

5. ACD BCD

1. Given

2. lines form rt.

5. SAS SAS

Given: FQ bisects AS

Prove: FAT QST

Statement 1. FQ bisects AS

A S Angle

2. AT ST Side

3. ATF & STQ are vertical

4. ATF STQ Angle

5. FAT QST

Reasons 1. Given

4. All vert. are

5. ASA ASA

Given: A D and BCA FED

AEF BCD

Prove: ABC DFE

Statement

1. A D Angle and

BCA FED Angle

Reasons 1. Given

AE CD and AEF BCD

2. EC EC

3. AE + EC CD + EC or

AC DE Side

4. ABC DFE

2. Reflexive Post. 3. Addition Post.

4. ASA ASA

Given: SU QR, PS RT

TSU QRP

Prove: PQR STU

Statement

1. SU QR, PS RT

TSU QRP

2. SR SR 3. PS + SR = RT + SR or

4. PQR STU

5. Q U

Reasons 1. Given 2. Reflexive Post. 3. Addition Post

4. SAS SAS

Given: M D

THE SEM

Prove: MTH DSE

Statement

1. M D Angle, ME HD

THE SEM

2. HE HE

3. ME – HE HD - HE or

MH DE Side

4. THM SED Angle

5. MTH DSE

Reasons 1. Given 2. Reflexive post. 3. Subtraction Post.

4. Supplements of are

5. ASA ASA

Given; SQ bisects PSR

Prove: PQS QSR

Statement

1. SQ bisects PSR

P R Angle

2. PSQ RSQ Angle

3. SQ SQ Side

4. PQS QSR

Reasons 1. Given

2. an bisectors cuts an into 2 parts. 3. Reflexive Post

4. AAS AAS

#1 #65

Given: PQ QS and TS QS R midpoint of QS

Prove: P T

Statement

1. PQ QS and TS QS R midpoint of QS

2. Q and S are rt.

3. Q S Angle

4. PRQ and TRS are vertical

5. PRQ TRS Angle

6. QR SQ Side

Reasons 1. Given

2. lines form rt.

3. All rt. are

5. All vert. are

6. A midpoint cuts a segment into 2

7. PQR TSR

8. P T

parts.

7. ASA ASA

Given: CB FB, BT BV

DV TS, DC FS

Prove: D S

Statement

1. CB FB, BT BV

DV TS, DC FS Side

2. BTV BVT Angle

3. CB + BT FB + BV or

CT FV Side

4. VT VT

5. DV + VT TS + VT or

DT SV Side

6. DCT SVF

7. D S

Reasons 1. Given

2. opp. sides in a are 3. Addition Post 4. Reflexive Post. 5. Addition Post

6. SAS SAS

#1 #67

Given: PQ DE and PB AE

QA PE and DB PE

Prove: D Q

Statement

1. PQ DE Hyp and PB AE

QA PE and DB PE

2. AB AB 3. PB – AB = AE – AB or

PA EB Leg

4. QAP and DBA are rt.

Reasons 1. Given 2. Reflexive post. 3. Subtraction Post.

4. lines form rt.

5. QAP DBA

6. PAQ EBD

7. D Q

5. All rt. are

6. HL HL

Given: TS TR

Prove: PS QR

Statement

1. TS TR Side

P Q Angle

2. PTS and QTR are vertical

3. PTS QTR Angle

4. PTS QTR

5. PS QR

Reasons 1. Given

3. All vert. are

4. AAS AAS

Given: HY and EV bisect each other

Prove: HE VY

Statement 1. HY and EV bisect each other

2. HA YA Side and EA VA Side

3. HAE and YAV are vertical

4. HAE YAV Angle

5. HAE YAV

6. HE VY

Reasons 1. Given

4. All vert. are

5. SAS SAS

#1 #70

Given: E D and A C B is the midpoint of AC

Prove: EA DC

Statement

1. E D Angle and A C Angle B is the midpoint of AC

2. EA DC Side

3. ABE CBE

4. EA DC

Reasons 1. Given

3. AAS AAS

Given: E is midpoint of AB

DA AB and CB AB

Prove: AD CB

Statement 1. E is midpoint of AB

DA AB and CB AB

2. AE EB Side

3. DE CE Side

Reasons 1. Given

3. opp. sides in a are

4. ADE BCD

5. AD CB

4. HL HL

Given: ABC CD bisects AB CD AB Prove: ACD BCDmwebster23.weebly.com/uploads/5/5/2/9/55298051/...#1...

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Honors Geometry Chapter 2 Review!. Name the property illustrated below. If segment AB is congruent to segment CD, then AB=CD. A.) definition of a midpoint

CD-Liste Dirigenten ABC Name ab: Carlo Sabajno 19.02 S.pdf · CD-Liste Dirigenten ABC... Name ab: Carlo Sabajno 19.02.12 Aureliano Pertile, Carena, Irene Minghini-Cattaneo, Apollo

Exercise 15 - cdn1.byjus.com · AB = 8 cm CD = 6 cm OM and ON are perpendiculars from O to AB and CD respectively. M and N are the Mid-points of AB and CD respectively In figure (i)

Geometry Regents Exam 0114 · ... Plane R is parallel to plane S. 2) ... F are points on AB and CD, respectively, BE ... diameter AB bisects chord

AB + CD AD + CB

h u h 20 3 20 1.732 36 - KopyKitab · q qAOB and COD60 60 In 'AOB, tan60 AB BO q 3 3 AB BO AB BO Also, in 'COD, tan30 1 3 3 CD DO CD DO DO CD q As, BD 80 80 3 80 3 BO DO AB CD 3 80

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Measuring Angles. Geometry vs Algebra Segments are Congruent –Symbol [ ] –AB CD – 1 2 Lengths of segments are equal. –Symbol [ = ] –AB = CD

n ab > cd ab < cd ab cd n ab n k u x n k - ViitoriOlimpici.ro · 2017-04-18 · Concursul Gazeta Matematic i ViitoriOlimpici.ro Concursul Gazeta Matematic i ViitoriOlimpici.ro

Question Bank In Mathematics Class IX (Term II) · Or,AC bisects C Proved. (ii) In ABC and CDA AC = AC [Common] BAC = DAC [Given] ... ABC ADC [ASA congruence] AB = AD and CB = CD

AB=CD Pattern

AB CD Wer wird Millionär? AB CD Wie ist der Fisch? in der Schuleklug leicht gemein

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Tel: 301-922-9508 Email: chiefmathtutor@gmail · 2017. 12. 5. · THE *Education Center AMC 10 2001 In trapezoid ABCD, AB and CD are perpendicular to AD, with AB+CD = BC, AB < CD,

Dynamic A/B testing with AB/CD

AB CD EF GH I J

May 2011 CHEMISTRY hours · May 2011 . CHEMISTRY 2Y2 hours . ... AB + CD ---+ AC +BD D. AB+CD -.. AD+CD . 12. A metal oxide in which oxygen has the

AB CD = BC FirstRanker

Lynn Creek bisects property - UNL Water Bioswales.pdf · Lynn Creek bisects property ... North Bioswales ~5 ac 2 ac impervious, 45% ... Microsoft PowerPoint - Krohn_NDOR Bioswales

Given: ABC CD bisects AB CD AB Prove: ACD BCD · #1 #5 Given: AEB & CED intersect at E E is the midpoint AEB AC ... 5. Intersect lines form vertical 6. Vertical are 7. ASA ASA #7