About congruent triangles

In school, we all learned the conditions for determining the congruence of triangles. There are usually 5 kinds of things mentioned in the textbook : SSS,SAS,AAS,ASA,HLSSS,SAS,AAS,ASA,HL.We notice that the above conditions are all around angles and edges, but there are other lines and points, such as Angle bisectors, heights, centers, etc. Can we use other conditions to determine the congruence of triangles? I made three proofs by myself. But there are too many of these to do it myself. If anyone has a great idea, let me know in the comments section and I'll add it to my note.

When two triangle have two equal edges and one side has an equal midian,they are congruent figures.


In the picture, AC=EGAC=EG,BC=FGBC=FG, ADAD and EHEH are midians on BCBC and FGFG.

D\because D and HH are midpoint on BCBC and FGFG,there are BC=FGBC=FG.     CD=GH,BD=FH\implies CD=GH,BD=FH.

In the ACD\triangle ACD and EGH\triangle EGH, AD=EH,AC=EG,GH=CDAD=EH,AC=EG,GH=CD

ACDEGH(SSS)    ACB=EGF\therefore \triangle ACD \cong \triangle EGH(SSS)\implies\angle ACB=\angle EGF

In the ABC\triangle ABC and EFG\triangle EFG, AC=EG,ACB=EGF,BC=FGAC=EG,\angle ACB=\angle EGF,BC=FG

ABCEFG(SAS)\therefore \triangle ABC\cong \triangle EFG(SAS)

When two triangles have two equal edges and the other side has an equal midian, they are congruent figures.


In the picture, AC=DFAC=DF, BC=EFBC=EF, CGCG and FHFH are midians on ABAB and DEDE

We extend CGCG and FHFH to II and JJ, and let CG=IJ=12CI,FH=JH=12FJCG=IJ=\dfrac12 CI, FH=JH=\dfrac12 FJ.

    G\implies G and HH are midpoints on CICI and FGFG. Also, GG and HH are midpoint on ABAB and DEDE.

    AIBC\implies AIBC and DJEFDJEF are parallelograms.     AC=BI,DF=EJ    BI=EJ\implies AC=BI, DF=EJ\implies BI=EJ

AC=DF,BC=EF,CG=FH    2CG=2FH    CI=FJ\because AC=DF, BC=EF, CG=FH \implies 2CG=2FH \implies CI=FJ

In the BCI\triangle BCI and EFJ\triangle EFJ, BC=EF,CI=FJ,BI=EJBC=EF, CI=FJ, BI=EJ

BCIEFJ(SSS)    EFJ=BCI\therefore \triangle BCI \cong \triangle EFJ(SSS) \implies \angle EFJ=\angle BCI

Also, we can know ACIDFJ(SSS)    ACI=DFJ\triangle ACI\cong \triangle DFJ(SSS) \implies \angle ACI=\angle DFJ

ACI+BCI=DFJ+EFJ    ACB=DFE\therefore \angle ACI+\angle BCI=\angle DFJ+\angle EFJ\implies \angle ACB=\angle DFE

In the ABC\triangle ABC and DEF\triangle DEF,AC=DF,ACB=DFE,BC=EFAC=DF, \angle ACB=\angle DFE, BC=EF

ABCDEF(SAS)\therefore \triangle ABC\cong \triangle DEF(SAS)

When two triangles have two equal angles and they clip edge has an equal midian, they're congruent figures.


In the picture, B=E,C=F,AH\angle B=\angle E, \angle C=\angle F,AH and DGDG are midians on BCBC and EFEF

In the ABC\triangle ABC and DEF\triangle DEF, B=E,C=F\angle B=\angle E, \angle C=\angle F

ABCDEF(AA)    ACDF=ABDE=BCEF\therefore \triangle ABC \sim \triangle DEF (AA) \implies \dfrac{AC}{DF}=\dfrac{AB}{DE}=\dfrac{BC}{EF}

AH=DG    AHDG=BCEF=1    BC=EF\because AH=DG \implies \dfrac{AH}{DG}=\dfrac{BC}{EF}=1 \implies BC=EF

In the ABC\triangle ABC and DEF\triangle DEF, B=E,C=F,BC=EF\angle B=\angle E, \angle C=\angle F, BC=EF

ABCDEF(ASA)\therefore \triangle ABC\cong \triangle DEF(ASA)

Note by Edward Christian
3 weeks, 5 days ago

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