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

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(D) In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$.Since $\overline{MN} \parallel \overline{BC}$,$\angle AMN \cong \angle

Vedclass · Accepted Answer

$7.5$

Vedclass · Answer

(D) In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$.Since $\overline{MN} \parallel \overline{BC}$,$\angle AMN \cong \angle ABC$ and $\angle ANM \cong \angle ACB$ (Corresponding angles).By $AA$ similarity criterion,$\Delta AMN \sim \Delta ABC$.Therefore,the ratio of corresponding sides is equal:$\frac{AM}{AB} = \frac{MN}{BC}$We know that $AB = AM + MB = 2 + 3 = 5$.Substituting the values:$\frac{2}{5} = \frac{3}{BC}$$2 	imes BC = 3 	imes 5$$2 	imes BC = 15$$BC = \frac{15}{2} = 7.5$

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $AM = 2$,$MB = 3$ and $MN = 3$,find $BC$.

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