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

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

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$2.8$

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(B) Given that in $\Delta ABC$,$\overline{MN} \parallel \overline{BC}$.According to the Basic Proportionality Theorem (Thales Theorem),if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points,the other two sides are divided in the same ratio.Therefore,$\frac{AM}{MB} = \frac{AN}{NC}$.First,find $AM$: Since $A-M-B$,$AM = AB - MB = 10.8 - 4.5 = 6.3$.Let $AN = x$. Then $NC = AC - AN = 4.8 - x$.Substituting the values into the ratio: $\frac{6.3}{4.5} = \frac{x}{4.8 - x}$.Simplify the ratio $\frac{6.3}{4.5} = \frac{63}{45} = \frac{7}{5} = 1.4$.So,$1.4 = \frac{x}{4.8 - x}$.$1.4(4.8 - x) = x$.$6.72 - 1.4x = x$.$6.72 = 2.4x$.$x = \frac{6.72}{2.4} = 2.8$.Thus,$AN = 2.8$.

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AB = 10.8$,$MB = 4.5$ and $AC = 4.8$,find $AN$.

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