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

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(C) 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 s

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$2$

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(C) 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,then the other two sides are divided in the same ratio.Given $\overline{MN} \parallel \overline{BC}$,therefore $\frac{AM}{MB} = \frac{AN}{NC}$.We are given $\frac{AM}{MB} = \frac{4}{5}$ and $NC = 2.5$.Substituting these values into the equation: $\frac{4}{5} = \frac{AN}{2.5}$.Multiplying both sides by $2.5$: $AN = \frac{4}{5} 	imes 2.5$.$AN = 4 	imes 0.5 = 2$.Thus,the value of $AN$ is $2$.

Part $I$	Part $II$
$1.$ In $\Delta ABC$ and $\Delta PQR, \angle A \cong \angle P$ and $\angle C \cong \angle Q$	$a.$ Correspondence $ABC \leftrightarrow RQP$ is a similarity.
$2.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{QR} = \frac{BC}{PQ}$ and $\angle B \cong \angle Q$	$b.$ Correspondence $ABC \leftrightarrow QPR$ is a similarity.
$3.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{BC}{PR} = \frac{CA}{QR}$	$c.$ Correspondence $ABC \leftrightarrow PQR$ is a similarity.
$4.$ In $\Delta ABC$ and $\Delta PQR, \frac{AB}{PQ} = \frac{CA}{PR}$ and $\angle A \cong \angle P$	$d.$ Correspondence $ABC \leftrightarrow PRQ$ is a similarity.

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $\frac{AM}{MB} = \frac{4}{5}$ and $NC = 2.5$,find $AN$.

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