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#1Given: ABCCD bisects ABCD ABProve: ACD BCDStatement ABCCD bisects ABCD AB2. AD DB SideReasons1.1. Given3.4.5.6.2. A bisector cuts a segment into 2 parts.3. lines form right .4. All rt are .5. Reflexive post.6. SAS SAS CDA and CDB are right CDA CDB AngleCD CD Side ACD BCD#2Given: ABC and DBE bisect eachother.Prove: ABD CBD1.2.3.4.5.StatementABC and DBE bisect each other.AB BC SideBD BE SideABD and BEC are vertical ABD BEC Angle ABD CBDReasons1. Given2. A bisector cuts a segment into 2 parts.3. Intersecting lines form vertical .4. Vertical are .5. SAS SAS

#1#3Given: AB CD and BC DA DAB, ABC, BCD and CDAare rt Prove: ABC ADC1.2.3.4.StatementAB CD SideBC DA Side DAB, ABC, BCD and CDAare rt ABC ADC Angle ABC ADCReasons1. Given2. Given3. All rt are .4. SAS SAS#4Given: PQR RQSPQ QSProve: PQR RQSStatement1. PQR RQS AnglePQ QSSide2. RQ RQSide3. PQR RQSReasons1. Given2. Reflexive Post.3. SAS SAS

#1#5Given: AEB & CED intersect at EE is the midpoint AEBAC AE & BD BEProve: AEC BEDStatement1. AEB & CED intersect at EE is the midpoint AEBAC AE & BD BE2. AEC and BED are vertical3. AEC BED Angle4. AE EBSide5. A & B are rt. 6. A B Angle7. AEC BEDReasons1. Given2. Intersecting lines form vertical .3. Vertical are .4. A midpoint cut a segment into 2 parts5. lines form right .6. All rt are .7. ASA ASA#6Given: AEB bisects CEDAC CED & BD CEDProve: EAC EBDStatement1. AEB bisects CEDAC CED & BD CED2. CE ED Side3. ACE & EDB are rt 4. ACE EDBAngleReasons1. Given2. A bisector cuts an angle into2 parts.3. Lines form rt .4. All rt are

#15. AEC & DEB are vertical 6. AEC DEBAngle7. EAC EBD#75. Intersect lines form vertical 6. Vertical are 7. ASA ASAGiven: ABC is equilateralD midpoint of ABProve: ACD BCDStatement1. ABC is equilateralD midpoint of AB2. AC BC Side3. AD DB Side4. CD CDSide5. ACD BCDReasons1. Given2. All sides of an equilateral are 3. A midpoint cuts a segment into2 parts.4. Reflexive Post5. SSS SSS#8Given: m A 50, m B 45,AB 10cm, m D 50m E 45 and DE 10cmProve: ABC DEFStatement1. m A 50, m B 45,AB 10cm, m D 50m E 45 and DE 10cm2. A D Angle and B E AngleAB DE Side3. ABC DEFReasons1. Given2. Transitive Prop3. ASA ASA

#1#9Given: GEH bisects DEFm D m FProve: GFE DEHStatement1. GEH bisects DEFm D m F Angle2. DE EFSide3. 1 & 2 are vertical4. 1 2Angle5. GFE DEHReasons1. Given2. Bisector cut a segment into 2 parts.3. Intersect lines form vertical 4. Vertical are 5. ASA ASA#10Given: PQ bisects RS at M R SProve: RMQ SMPStatement1. PQ bisects RS at M R S Angle2. RM MS SideReasons1. Given2. Bisector cut a segment into 2

#13. 1 & 2 are vertical angles4. 1 2Angle5. RMQ SMPparts3. Intersect lines form vertical 4. Vertical are 5. ASA ASA#11Given: DE DGEF GFProve: DEF DFGStatement1. DE DG SideEF GF Side2. DF DF Side3. DEF DFGReasons1. Given2. Reflexive Post3. SSS SSS#12Given: KM bisects LKJLK JKProve: JKM LKMStatement1. KM bisects LKJLK JK Side2. 1 2AngleReasons1. Given2. An bisectors cuts the into2 parts

#13. KM KM Side4. JKM LKM3. Reflexive Post4. SAS SAS#13Given: . PR QR P QRS is a medianProve: PSR QSRStatement1. PR QRSide P QAngleRS is a medianSide2. PS SQ3. PSR QSRReasons1. Given2. A median cuts the side into2 parts3. SAS SAS#14Given: EG is bisectorEG is an altitudeProve: DEG GEFStatement1. EG is bisectorEG is an altitude2. 3 4 AngleReasons1. Given2. An bisector cuts an into2 parts.

#13. EG DF4. 1 & 2 are rt 5. 1 2Angle6. GE GESide7. DEG GEF3.4.5.6.7.An altitude form lines. lines form right angles.All right angles are Reflexive PostASA ASA#15Given:AEABProve:1.2.3.4.5.6.Statement A and D are a rt AE DFSideAB CD A D AngleBC BCAB BC CD BCor AC BDSide AEC DFBEC FB A and D are a rt DF CDEC FBReasons1. Given2. All right angles are .3. Reflexive Post.4. Addition Prop.5. SAS SAS6. Corresponding parts of are .#16Given: CA CBD midpoint of ABProve: A BStatement1. CA CB SideD midpoint of ABReasons1. Given

#12. AD DBSide3. CD CDSide4. ADC DBC5. A B2. A midpoint cuts a segment into2 parts3. Reflexive Post4. SSS SSS5. Corresponding parts of are .#17Given: . AB CD CAB ACDProve: AD CB1.2.3.4.StatementAB CD Side CAB ACDAngleAC ACSide ACD ABCAD CBReasons1. Given2. Reflexive Post3. SAS SAS4. Corresponding parts of are .#18Given: AEB & CED bisect eachOtherProve: C DStatement1. AEB & CED bisect each other2. CE ED Side & AE EB Side3. 1 and 2 are verticalReasons1. Given2. A bisector cuts segments into2 parts.3. Intersect lines form vertical

#14. 1 2 Angle5. AEC DEB6. C D4. Vertical are 5. SAS SAS6. Corresponding parts of are#19Given: KLM & NML are rt KL NMProve: K N1.2.3.4.5.Statement KLM & NML are rt KL NMSide KLM NML AngleLM LMSide KLM LNM K NReasons1. Given2.3.4.5.All rt are Reflexive Post.SAS SASCorresponding parts of are .#20Given: AB BCPA PDProve: a) APBb) APCStatement1. AB BC CD SidePA PD Side & PB PC Side2. ABP CDP3. APB DPC CD& PB PC DPC DPBReasons1. Given2. SSS SSS3. Corresponding parts of are .

#14. BPC BPC5. APB BPC DPC BPCor APC DPB4. Reflexive Post.5. Addition Prop.#21Given: PM is AltitudePM is medianProve: a) LNP is isoscelesb) PM is bisector1.2.3.4.5.StatementPM is Altitude & PM is medianPM LN 1 and 2 are rt 1 2LM MN6. PM PM7. LMP PMN8. PL PN9. LNP is isosceles10. LPN MPN11. PM is bisector#22Reasons1. Given2. An altitude form lines.3. lines form right angles.4. All right angles are 5. A median cuts the side into2 parts6. Reflexive Post.7. SAS SAS8. Corresponding parts of are .9. An Isosceles is a with2 sides10.Corresponding parts of are .11. A bisector cuts an into2 parts

#1Given: CA CBProve: CAD CBEStatement1. CA CB2. 2 33. 1 & 2 are supplementary 3 & 4 are supplementary4. 1 4 or CAD CBEReasons1. Given2. If 2 sides are then the oppositeare .3. Supplementary are form by alinear pair.4. Supplement of are .#23Given: AB CB & AD CDProve: BAD BCDStatement1. AB CB & AD CD2. 1 2 3 43. 1 3 2 4or BAD BCD#24Reasons1. Given2. If 2 sides are then the oppositeare .3. Addition Post.

#1Given: ΔABC ΔDEFM is midpoint of ABN is midpoint DEProve: ΔAMC ΔDNF1.2.3.4.StatementΔABC ΔDEFM is midpoint of ABN is midpoint DE D A Angle and DF AC SideAM MB and DN NE Side5. ΔAMC ΔDNFReasons1. Given2. Given3. Corresponding parts of Δ are 4. A midpoint cuts a segment into2 parts5. SAS SAS#25Given: ΔABC ΔDEFCG bisects ACBFH bisects DFEProve: CG FHStatement1. ΔABC ΔDEFCG bisects ACBFH bisects DFEReasons

#1#26Given: ΔAME ΔBMFDE CFProve: AD BC1.2.3.4.5.StatementΔAME ΔBMFDE CFEM MFAM MB Side 1 2 AngleDE EM CF MFor DM MC SideΔADM ΔBCMAD BCReasons1. Given2. Corresponding parts of Δ are 3. Addition Post.4. SAS SAS5. Corresponding parts of Δ are Given: AEC & DEB bisect eachotherProve: E is midpoint of FEGStatement1. AEC & DEB bisect each otherReasons1. Given

#12. DE BE Side and AE EC Side3. AEB & DEC are vertical 4. AEB DEC Angle5. ΔAEB ΔDEC6. D B7. 1 & 2 are vertical angles8. 1 29. ΔGEB ΔDEF10. GE FE11. E is midpoint of FEG2. A bisector cuts a segment into2 parts.3. Intersecting lines form vertical 4. Vertical are .5. SAS SAS6. Corresponding parts of Δ are 7. Intersecting lines form vertical 8. Vertical are .9. ASA ASA10. Corresponding parts of Δ are 11. A midpoint divides a segmentinto 2 parts.#28Given: BC BABD bisects CBAProve: DB bisects CDAStatementReasons

#11. BC BA SideBD bisects CBA2. 1 2 Angle3.4.5.6.BD BDSideΔABD ΔBCD 3 4DB bisects CDA1. Given2. A bisector cuts an angle into2 parts.3. Reflexive Post.4. SAS SAS5. Corresponding parts of Δ are 6. A angle bisector cuts an angleinto 2 parts.#29Given: AE FBDA CB A and B are Rt. Prove: ADF CBEDF CEStatement1. AE FBDA CB Side A and B are Rt. 2. EF EF3. AE EF FB EFor AF EB SideReasons1. Given2. Reflexive Post3. Addition Property

#14. A B Angle5. ADF CBE6. DF CE4. All rt. are .5. SAS SAS6. Corresponding parts of Δ are #30Given: SPR SQTPR QTProve: SRQ STP R T1.2.3.4.5.StatementSPR SQT SidePR QT S S AngleSPR – PR SQT – QTor SR ST Side SRQ STP R TReasons1. Given2. Reflexive Post3. Subtraction Property4. SAS SAS5. Corresponding parts of Δ are #31Given: DA CBDA AB & CB ABProve: DAB CBAAC BD1.2.3.4.5.6.StatementDA CB SideDA AB & CB AB DAB and CBA are rt DAB CBA AngleAB AB Side DAB CBAAC BDReasons1. Given2.3.4.5.6. lines form rt .All rt are .Reflexive post.SAS SASCorresponding parts of Δ are .

#1#32Given: BAE CBF BCE CDFAB CDProve: AE BF E F1.2.3.4.5.Statement BAE CBF Angle BCE CDF AngleAB CDBC BCAB BC CD BCor AC BD Side AEC BDFAE BF E FReasons1. Given2. Reflexive Post.3. Addition Property.4. ASA ASA5. Corresponding parts of Δ are .#33Given: TM TNM is midpoint TRN is midpoint TSProve: RN SMStatementReasons

#11. TM TN SideM is midpoint TRN is midpoint TS2. T T Angle3. RM is ½ of TRNS is ½ of TS4. RM NS5. TM RM TN NSor RT TS Side6. RTN MTS7. RN SM1. Given2. Reflexive Post.3. A midpoint cuts a segment in .4. ½ of parts are .5. Addition Property6. SAS SAS7. Corresponding parts of Δ are .#34Given: AD CE & DB EBProve: ADC CEAStatement1. AD CE & DB EB SideReasons1. Given

#12. B B Angle3. AD DB CE EBor AB BC Side4. ABE BCD5. 1 26. 1 & 3 are supplementary 2 & 4 are supplementary7. 3 4 or ADC CEA2. Reflexive Post3. Addition Post.4. SAS SAS5. Corresponding parts of Δ are .6. A st. line forms supplementary .7. Supplements of are .#35Given: AE BF & AB CD ABF is the suppl. of AProve: AEC BFDStatement1. AE BF Side & AB CD ABF is the suppl. of AReasons1. Given

#12. A 1 Angle3. BC BC4. AB BC CD BCor AC BD Side5. AEC BFD2. Supplements of are .3. Reflexive Post.4. Addition Property.5. SAS SAS#36Given: AB CBBD bisects ABCProve: AE CEStatement1. AB CB SideBD bisects ABC2. 1 2Angle3. BE BE Side4. BEC BEA5. AE CEReasons1. Given2. A bisector cuts an into2 parts.3. Reflexive Post.4. SAS SAS5. Corresponding parts of Δ are #37Given: PB PCProve: ABP DCPStatement1. PB PCReasons1. Given

#12. 1 23. 1 & ABP are supplementary 2 & DCP are supplementary 4. ABP DCP2. opposite sides are .3. Supplementay are formed by alinear pair.4. Supplements of are .#38Given: AC and BD are bisectors ofeach other.Prove: AB BC CD DA1.2.3.4.StatementAC and BD are bisectors ofeach other 1, 2, 3 and 4 are rt 1 2 3 4 AngleAE EC and BE DE 2 sides5. ABE BEC DEC AED6. AB BC CD DAReasons1. Given2. lines form rt .3. All rt are .4. A bisector cuts a segment into2 parts.5. SAS SAS6. Corresponding parts of Δ are #39Given: AEFB, 1 2CE DF, AE BFProve: AFD BECStatementReasons

#11. AEFB, 1 2 AngleCE DF Side, AE BF2. EF EF3. AE EF BF EF orAF EB Side4. AFD BEC1. Given2. Reflexive Post.3. Addition Property4. SAS SAS#40Given: SX SY, XR YTProve: RSY TSX1.2.3.4.StatementSX SY Side, XR YTSX XR SY YTor SR ST Side S SAngle RSY TSXReasons1. Given2. Addition Post.3. Reflexive Post.4. SAS SAS#41Given: DA CBDA AB, CB ABProve: DAB CBA

#11.2.3.4.5.StatementDA CB SideDA AB, CB AB DAB and CBA are rt. DAB CBA AngleAB ABSide DAB CBAReasons1. Given2.3.4.5. lines form rt All rt. are Reflexive Post.SAS SAS#42Given: AF EC 1 2, 3 4Prove: ABE CDFStatement1. AF EC 1 2, 3 4 Angle2. DFC BEAAngle3. EF EF4. AF EF EC EF orAE FC Side5. ABE CDF#43Reasons1. Given2. Supplements of are 3. Reflexive post.4. Addition Post.5. AAS AAS

#1Given: AB BF, CD BF 1 2, BD FEProve: ABE CDF1.2.3.4.5.6.StatementAB BF, CD BF 1 2 Side , BD FE B and CDF are rt. B CDF AngleDE DEBD DE FE DE orBE DF Side ABE CDFReasons1. Given2.3.4.5. lines form rt. All rt. are Reflexive Post.Addition Post.6, ASA ASA#44Given: BAC BCACD bisects BCAAE bisects BACProve: ADC CEA1.2.3.4.5.Statement BAC BCA AngleCD bisects BCAAE bisects BAC ECA ½ BAC and DCA ½ BCA ECA DCA AngleAC ACSide ADC CEAReasons1. Given2. bisector cuts an in ½3. ½ of are 4. Reflexive post.5. ASA ASA

#1#45Given: TR TS, MR NSProve: RTN STMStatement1. TR TS Side, MR NS2, TR – MR TS – NS orTM TN Side3. T T Angle4. RTN STM#46Reasons1. Given2. Subtraction Post.3. Reflexive Post.4. ASA ASAGiven: CEA CDB, ABCAD and BE intersect at P PAB PBAProve: PE PDStatement1. CEA CDB, ABCAD and BE intersect at P PAB PBA2.Reasons1. Given

#1#47Given: AB AD and BC DCProve: 1 21.2.3.4.5.6.7.StatementAB AD and BC DCAC AC ABC ADCAE AE BAE DAE ABE ADE 1 2Reasons1.2.3.4.5.6.7.GivenReflexive Post.SSS SSSReflexive Post.Corresponding parts of Δ are .SAS SASCorresponding parts of Δ are .#48Given: BD is both median andaltitude to ACProve: BA BC1.2.3.4.5.6.StatementBD is both median andaltitude to ACAD CD Side ADB and CDB are rt. ADB CDB AngleBD BDSide ABD CBDReasons1. Given2. A median cuts a segment into 2 parts3. Line