中三-三角形[20點]

1.
a)
AB=DF(given)
EB=DC(given)
AC=EF(given)
CE=EC(given)
AC+CE=EF+CE
AE=CF
∴∆ABE≅∆FDC(SSS)
b)
∆ABE≅∆FDC(proved)
∴∠AEB=∠FCD(corr.∠s, ≅∆s)
∴DC//EB(alt.∠s eq.)
DC=EB(proved)
∴BCDE is a //gram.(opp. sides eq. and //)
∴BC=DE(opp. sides of //)

2.
a)
∠ACB=60°(equil. ∆)
CD=DE(given)
∴∠CDE=∠DEC(base ∠s, isos. ∆)
∠CDE+∠DEC=60°(ext. ∠s of ∆)
∴2∠DEC=60°
∠DEC=30°
b)
∆ABD≅∆CDB(prove by yourself)
hint: the reason is AAS/RHS
∴∠ABD=∠DBM
∠ABC=60°(equil. ∆)
∴∠DBM=60°/2=30°
∴∆BDM≅∆EDM(prove by yourself)
hint: the reason is AAS
∴BM=ME(corr. sides, ≅∆s)

If you really don't know how to prove the two pairs of congruent triangles, please ask me and I will list the steps.