mas ā” ABE = mas ā” CBE = 60 / 2 = 30 Ā°
mas ā” ACB = 180 - ( 90 + 60 ) = 30 Ā°
/\BCE isoscel cu BE = EC
mas ā” BEC = 180 - ( 30 + 30 ) = 120 Ā°
In /\ ABC , cu teoremei unghi de 30 Ā° vom rezulta :
=> AB = BC / 2 = 10 / 2 = 5
BC ^ 2 = AB ^ 2 + AC ^ 2
AC ^ 2 = BC ^ 2 - AB ^ 2
AC ^ 2 = 100 - 25
AC ^ 2 = 75
AC = radical 75 = 5 radical 3
In /\ ABE :
mas ā” B = 30 Ā°
mas ā” A = 90 Ā°
mas ā” E = 60 Ā°
Cu teorema ungh de 30 Ā°
AE = BE / 2 => BE = 2AE
BE = EC => 2AE = EC
dar AE + EC = AC = 5 radical 3
deci AE + 2AE = 5 radical 3
3AE = 5 radical 3 / 3 =>
=> EC = AC - AE = 5 radical 3 - 5 radical 3 / 3 = 10 radical 3 / 3
P/\BCE = BE + EC + BC = 10 radical 3 / 3 + 10 radical 3 / 3 + 10 = 10 ( 2 radical 3 + 3 ) / 3
