Question

1.Fie ABCD,un trapez,AB ll CD, iar AC intersectat BD={O}.Daca AB=10 cm,CD=25 cm,AC=35 cm si BD=42 cm,aflati lungimile segmentelor:[AO],[CO],[BO],[DO].

2.In triunghiul ABC consideram punctul M ∈(AB) si MN ll AC,N ∈ (BC),MP ll BC,P ∈(AC).Demonstrati ca:$\frac{MN}{AC} + \frac{MP}{BC} =1$

Answer 1

1) Triunghiurile AOB si COD sunt asemenea conform TFA (AB II CD) => AO/CO=BO/DO=AB/DC=10/25=2/5
AO/CO=2/5 =>AO/AC=2/7 => AO=2*35/7=10 cm=> CO=25 cm
BO/DO=2/5 => BO/BD=2/7=> BO=2*42/7=12 cm=> DO=30 cm

2) Triunghiurile BMN si BAC sunt asemenea conform TFA (ACIIMN)
=> BM/AB=MN/AC
Analog: Triunghiurile AMP si ABC sunt asemenea => AM/AB=MP/BC.
 Cand le adunam ne da exact 1.