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Prove that a line drawn parallel to one side of a triangle to intersect the other two sides in distinct points divides the other two sides in the same ratio. Hence, in the figure given below, prove that $\frac{AM}{MB} = \frac{AN}{ND}$ where LM $||$ CB and LN $||$ CD.
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Correct figure, given, to prove and construction
Correct proof
In $\triangle ABC$, LM $||$ CB
$\frac{AM}{MB} = \frac{AL}{LC}$ --- (1)
In $\triangle ADC$, LN $||$ CD
$\frac{AN}{ND} = \frac{AL}{LC}$ --- (2)
from (1) and (2), we have
$\frac{AM}{MB} = \frac{AN}{ND}$
Correct proof
In $\triangle ABC$, LM $||$ CB
$\frac{AM}{MB} = \frac{AL}{LC}$ --- (1)
In $\triangle ADC$, LN $||$ CD
$\frac{AN}{ND} = \frac{AL}{LC}$ --- (2)
from (1) and (2), we have
$\frac{AM}{MB} = \frac{AN}{ND}$