1. m(<DBC)=m(<DBA)+m(<ABC). Dar m(<DBA)=m(<ACE)=60* si m(<ABC)=m(<ACB) => m(<DBC)=m(<ECB).
ΔABD-echilateral => [BD]≡[AB].
ΔACE-echilateral => [CE]≡[AC].
Dar [AB]≡[AC], rezulta [BD]≡[AB].
[BC]≡[BC] } (LUL)
<DBC≡<ECB } => ΔDBC≡ΔECB => [CD]≡[BE] (1).
[BD]≡[CE] }
m(<DBC)=m(<DBA)+m(<ABC).
m(<ABF)=m(<CBF)+m(<ABC),
Dar m(<DBA)=m(CBF)=60* => m(<DBC)=m(<ABF).
[BD]≡[AB] } (LUL)
<DBC≡<ABF } => ΔBCD≡ΔBFA => [CD]≡[AF] (2).
[BC]≡[BF] (echilateral) }
Din (1) si (2) => [AF]≡[CD]≡[BE].
2.
[tex] a^{2}=bc~\Rightarrow abc= a^{3}. \\ b^{2} =ac ~\Rightarrow abc= b^{3}. \\ Rezulta~ca~ a^{3}= b^{3} \Rightarrow a=b. \\ a^{2}=bc \Rightarrow c= \frac{ a^{2} }{b} ,~dar~a=b~\Rightarrow~c= \frac{ b^{2} }{b}=b. \\ \\ Avem~a=b~si~b=c,~rezulta~\boxed{a=b=c}~(adica~triunghiul~este~ \\ ECHILATERAL)[/tex]
Desenul de la prima problema ([AB]≡[AC]) este mai jos:
