Răspuns :
a) DM || AB =>(Thales) Δ ABR ~ Δ DMR => [tex] \frac{AB}{DM} = \frac{BR}{MR} = \frac{AR}{DR} [/tex] => [tex] \frac{AB}{DM} = \frac{AR}{DR} [/tex] =>
=> [tex] \frac{12}{DC-MC} = \frac{AD+DR}{DR} [/tex] => [tex] \frac{12}{9} = \frac{6+RD}{RD} [/tex] <=>
<=> [tex] \frac{4}{3} = \frac{6+RD}{RD} [/tex] <=> 4RD = 3( 6+RD) => 4RD = 18 + 3RD => RD = 18 cm
c) [tex] \frac{AB}{DM} = \frac{BR}{MR} => \frac{4}{3} = \frac{BR}{MR} <=> 3BR = 4MR[/tex] <=> [tex]BR= \frac{4MR}{3} [/tex] (1)
Dar [tex] \frac{4-3}{3} = \frac{BR-MR}{MR} <=> \frac{1}{3} = \frac{BM}{MR} <=>BM = \frac{MR}{3} [/tex] (2)
Din (1) si (2) => BR = 4BM (3)
BN x BR = NR x BM <=> [tex] \frac{BR}{BM} = \frac{NR}{BN} [/tex] <=> [tex] \frac{NR}{BN} = 4 [/tex]
Deci mai trebuie sa aratam ca NR = 4BN si gata ... dar nu stiu cum. Sper ca te-am ajutat. Poti sa iei AB || MC =>(Thales).. ... => MP = 5 cm si CP = 2 cm, dar nu stiu daca te ajuta ...
=> [tex] \frac{12}{DC-MC} = \frac{AD+DR}{DR} [/tex] => [tex] \frac{12}{9} = \frac{6+RD}{RD} [/tex] <=>
<=> [tex] \frac{4}{3} = \frac{6+RD}{RD} [/tex] <=> 4RD = 3( 6+RD) => 4RD = 18 + 3RD => RD = 18 cm
c) [tex] \frac{AB}{DM} = \frac{BR}{MR} => \frac{4}{3} = \frac{BR}{MR} <=> 3BR = 4MR[/tex] <=> [tex]BR= \frac{4MR}{3} [/tex] (1)
Dar [tex] \frac{4-3}{3} = \frac{BR-MR}{MR} <=> \frac{1}{3} = \frac{BM}{MR} <=>BM = \frac{MR}{3} [/tex] (2)
Din (1) si (2) => BR = 4BM (3)
BN x BR = NR x BM <=> [tex] \frac{BR}{BM} = \frac{NR}{BN} [/tex] <=> [tex] \frac{NR}{BN} = 4 [/tex]
Deci mai trebuie sa aratam ca NR = 4BN si gata ... dar nu stiu cum. Sper ca te-am ajutat. Poti sa iei AB || MC =>(Thales).. ... => MP = 5 cm si CP = 2 cm, dar nu stiu daca te ajuta ...