In Delta ABC,A-M-B,A-N-C and overline{MN} par | vedclass

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(C) In $\Delta AMN$ and $\Delta ABC$,since $\overline{MN} \parallel \overline{BC}$,the triangles are similar by the $AA$ similarity criterion.Therefor

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$3:4$

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(C) In $\Delta AMN$ and $\Delta ABC$,since $\overline{MN} \parallel \overline{BC}$,the triangles are similar by the $AA$ similarity criterion.Therefore,$\frac{AM}{AB} = \frac{MN}{BC}$.Given $MN = 3$ and $BC = 7$,we have $\frac{AM}{AB} = \frac{3}{7}$.Since $AB = AM + MB$,we can write $\frac{AM}{AM + MB} = \frac{3}{7}$.By cross-multiplication,$7 AM = 3 AM + 3 MB$.$4 AM = 3 MB$.Therefore,$\frac{AM}{MB} = \frac{3}{4}$.Thus,$M$ divides $\overline{AB}$ from $A$ in the ratio $3:4$.

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $MN = 3$ and $BC = 7$,then $M$ divides $\overline{AB}$ from $A$ in the ratio.......

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