AB║DC, din teorema fundamentala a asemanarii⇒triunghiul MAB~MDC⇒
[tex] \frac{MA}{MD} = \frac{MB}{MC} = \frac{AB}{DC} [/tex]
Inlocuim AB=18, CD=24 si simplificam 18 si 24 prin 6.
Obtinem: [tex] \frac{MA}{MD} = \frac{MB}{MC} = \frac{3}{4}
[/tex]Facem proportii derivate: La numitori scadem numaratorii:
[tex] \frac{MA}{MD-MA} = \frac{MB}{MC-MB} = \frac{3}{4-3} [/tex]
Deci:
[tex] \frac{MA}{AD} = \frac{MB}{BC} = 3 [/tex]. Ilocuim pe AD=12 si BC=42.
Obtinem:
[tex] \frac{MA}{12} = \frac{MB}{42} = 3 [/tex]
De aici, MA=3*12=36
MB=3*42=126
MD=MA+AD=12+36=48
MC=MB+BC=126+42=168
