Prezentam o problema simpla rezolvata cu cazurile de asemanare

Stabiliti daca triunghiurile ABC si EFT sunt asemenea in urmatoarele situatii:

a) m\left(\widehat{B}\right)=65^{0}; m\left(\widehat{C}\right)=70^{0}; m\left(\widehat{E}\right)=45^{0}; m\left(\widehat{F}\right)=65^{0}

Demonstratie:

Cazurile de asemanarea a triunghiurilor
In triunghiul ABC stim

m\left(\widehat{B}\right)=65^{0}; m\left(\widehat{C}\right)=70^{0}
Deci putem afla masura unghiului A, astfel avem
m\left(\widehat{A}\right)+m\left(\widehat{B}\right)+m\left(\widehat{C}\right)=180^{0}\Rightarrow m\left(\widehat{A}\right)+65^{0}+70^{0}=180^{0}\Rightarrow m\left(\widehat{A}\right)=180^{0}-135^{0}\Rightarrow m\left(\widehat{A}\right)=45^{0}
In triunghiul EFT stim
m\left(\widehat{E}\right)=45^{0}; m\left(\widehat{F}\right)=65^{0}
Deci putem afla masura unghiului T
m\left(\widehat{E}\right)+m\left(\widehat{F}\right)+m\left(\widehat{T}\right)=180^{0}\Rightarrow m\left(\widehat{T}\right)+45^{0}+65^{0}=180^{0}\Rightarrow m\left(\widehat{T}\right)=180^{0}-110^{0}\Rightarrow m\left(\widehat{T}\right)=70^{0}
Deci avem:
Cazul de asemanare u.u.u
Astfel avem ca:

m\left(\widehat{ABC}\right)=m\left(\widehat{EFT}\right)=65^{0}  \\m\left(\widehat{ACB}\right)=m\left(\widehat{ETF}\right)=70^{0}
Dar si ca
m\left(\widehat{BAC}\right)=m\left(\widehat{FET}\right)=45^{0}.
Deci cu cazul de asemanare u.u, \Delta ABC\sim\Delta EFT.

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