Question

1)ΔABC
M∈(BC)
N∈(AC)
AB||MN
$\frac{AN}{NC} = 7\frac{1}{2}$
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[tex] \frac{MC}{BC} =?
[/tex]
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                                      SAU 2.

2)ΔABC:
D∈(AB)
E∈(AC)
AD=6
DB=2
AE=12
EC=3
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DE||BC
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Answer 1

1) $\frac{AN}{NC} =7 \frac{1}{2} = \frac{7*2+1}{2} = \frac{15}{2}$
deoarece AB||MN =>
$\frac{AN}{NC} = \frac{BM}{MC}$ ( adunam numaratorul la numitor)
$\frac{AN}{AN+NC} = \frac{BM}{BM+MC}$
$\frac{15}{15+2} = \frac{MC}{BC} \\ \frac{MC}{BC} = \frac{15}{17}$

2) pentru ca DE||BC trebuie ca urmatoarea conditie sa fie indeplinita:
$\frac{AD}{DB} = \frac{AE}{EC} \\ \\ \frac{6}{2}= \frac{12}{3} \\ 3=4 ( fals)$

Asadar DE nu este paralel cu BC