Punctul a)
[tex]AB=\dfrac{3}{5}MN\\ AC=0,6 MP=\dfrac{3}{5}MP\\ BC=60\% NP=\dfrac{60}{100}NP=\dfrac{3}{5}NP\\ \dfrac{AB}{MN}=\dfrac{AC}{MP}=\dfrac{BC}{NP} \stackrel{RTFA}\implies \tt \Delta ABC \sim \Delta MNP[/tex]
RTFA=Reciproca teoremei fundamentale ale asemănării.
Punctul b)
Trebuie ca ∡A=∡M, ∡B=∡N, ∡C=∡P
[tex]\measuredangle A=0,(6)\cdot 90^{\circ}=\dfrac{2}{3}\cdot 90^{\circ}=60^{\circ}\\ \measuredangle M=0,(3)\cdot 180^{\circ}=\dfrac{1}{3}\cdot 180^{\circ}=60^{\circ}[/tex]
[tex]\measuredangle N=180^{\circ}-60^{\circ}-80^{\circ}\\ \Rightarrow \measuredangle N=40^{\circ} \Rightarrow \measuredangle N \not= \measuredangle B(=70^{\circ})[/tex]
Unghiurile triunghiurilor nu sunt egale ⇒ triunghiurile ABC și MNP nu sunt asemenea.
Punctul c)
[tex]\dfrac{AB}{2}=\dfrac{MN}{3}\Rightarrow AB=\dfrac{2}{3}MN\\ \dfrac{BC}{10}=\dfrac{NP}{15}\Rightarrow BC=\dfrac{10}{15}NP=\dfrac{2}{3}NP[/tex]
Avem [tex]\tfrac{AB}{MN}=\tfrac{BC}{NP}[/tex] și ∡N=∡B [tex]\stackrel{L.U.L.}\implies \tt \Delta ABC \sim \Delta MNP[/tex]
Punctul d)
[tex]\dfrac{AB}{MN}=\dfrac{6}{3}=2\\ \dfrac{AC}{MP}=\dfrac{10}{2}=5\\ \Rightarrow \dfrac{AB}{MN}\not=\dfrac{AC}{MP} \Rightarrow \tt \Delta ABC \not\sim \Delta MNP[/tex]
Punctul e)
[tex]\dfrac{AB}{MN}=\dfrac{5}{2,5}=2\\ \dfrac{AC}{MP}=\dfrac{6,2}{3,1}=2\\ \dfrac{BC}{NP}=\dfrac{8}{4}=2\\ \dfrac{AB}{MN}=\dfrac{AC}{MP}=\dfrac{BC}{NP} \stackrel{RTFA}\implies \tt \Delta ABC \sim \Delta MNP[/tex]
