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

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(C) Given that $\overline{MN} \parallel \overline{BC}$ in $\Delta ABC$. According to the Basic Proportionality Theorem (Thales Theorem),if a line is p

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$1.6$

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(C) Given that $\overline{MN} \parallel \overline{BC}$ in $\Delta ABC$. According to the Basic Proportionality Theorem (Thales Theorem),if a line is parallel to one side of a triangle intersecting the other two sides,then it divides the two sides in the same ratio. Therefore,$\frac{AM}{AB} = \frac{AN}{AC}$. We are given $\frac{AM}{AB} = \frac{5}{7}$ and $AN = 4$. Substituting these values into the equation: $\frac{5}{7} = \frac{4}{AC}$. Cross-multiplying gives $5 	imes AC = 7 	imes 4$,so $5 	imes AC = 28$. $AC = \frac{28}{5} = 5.6$. Since $A-N-C$,we have $AC = AN + NC$. $5.6 = 4 + NC$. $NC = 5.6 - 4 = 1.6$.

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM : AB = 5 : 7$ and $AN = 4$,find $NC$.

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