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In figura alaturata  se stie ca ΔAMC≡ΔANB si [OB]≡[OC] Demonstrati
a)[BM];
b)ΔABC isoscel;
c)[OM]≡[ON]


In Figura Alaturata Se Stie Ca ΔAMCΔANB Si OBOC DemonstratiaBMbΔABC IsoscelcOMON class=

Răspuns :

Din tr AMC=tr ANB rezulta ca :
AB=AC (deci tr ABC este isoscel)  b)
AM=AN, deci si AB-AM=AC-AN, adica:
BM=CN  a)
m(<ABN)=m(<ACM)  (rel1)
m(<AMC)=m(<ANB), deci si
180-m(<AMC)=180-m(<ANB), adica:
m(<BMC)=m(<BNC)  (rel 2)
Din a), (rel1) si (rel2) rezulta ca tr OMB congruent cu tr ONC (U.L.U.), deci
OM=ON.



[tex]\bigtriangleup AMC \equiv \bigtriangleup ANB \\ \([OB] \equiv \([OC] \\ ---------- \\ \([BM]=\([CN] \\ \bigtriangleup ABC~isoscel \\ \([OM] \equiv \([ON] \\ ---------- \\ \bigtriangleup AMC \equiv \bigtriangleup ANB \Longrightarrow AM=AN \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~MC=NB \\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~AC=AB \\ a)BM=AB-AM \\ CN=AC-AN \\ AC=AB~si~AM=AN \Longrightarrow \boxed{BM=CN} \\ b)AB=AC \Longrightarrow \boxed{\bigtriangleup ABC ~isoscel} \\[/tex]
[tex]c)OM=MC-OC \\ ON=NB-OB \\ OB=OC~si~MC=NB \Longrightarrow \boxed{OM=ON}[/tex]