Step 1: Understanding the Concept:
The condition \(\frac{AB}{DE} = \frac{BC}{EF} = \frac{CA}{DF}\) means that the sides of \(\triangle ABC\) are proportional to the corresponding sides of \(\triangle DEF\). By the Side-Side-Side (SSS) similarity criterion, this implies that \(\triangle ABC \sim \triangle DEF\).
Step 2: Key Formula or Approach:
In similar triangles, corresponding angles are equal. The correspondence is given by the order of vertices in the ratio.
A corresponds to D, B corresponds to E, and C corresponds to F.
Therefore, \(\angle A = \angle D\), \(\angle B = \angle E\), and \(\angle C = \angle F\).
The sum of angles in a triangle is 180°.
Step 3: Detailed Explanation:
We need to find the measure of \(\angle F\). Since \(\triangle ABC \sim \triangle DEF\), we have \(\angle F = \angle C\).
We can find \(\angle C\) using the angle sum property in \(\triangle ABC\).
We are given \(\angle A = 40^\circ\) and \(\angle B = 80^\circ\).
\[ \angle A + \angle B + \angle C = 180^\circ \]
\[ 40^\circ + 80^\circ + \angle C = 180^\circ \]
\[ 120^\circ + \angle C = 180^\circ \]
\[ \angle C = 180^\circ - 120^\circ = 60^\circ \]
Since \(\angle F = \angle C\), the measure of \(\angle F\) is 60°.
Step 4: Final Answer:
The measure of \(\angle F\) is 60°.
In the following figure \(\angle\)MNP = 90\(^\circ\), seg NQ \(\perp\) seg MP, MQ = 9, QP = 4, find NQ. 
Solve the following sub-questions (any four): In \( \triangle ABC \), \( DE \parallel BC \). If \( DB = 5.4 \, \text{cm} \), \( AD = 1.8 \, \text{cm} \), \( EC = 7.2 \, \text{cm} \), then find \( AE \). 
In the following figure, XY \(||\) seg AC. If 2AX = 3BX and XY = 9. Complete the activity to find the value of AC.
Activity:
2AX = 3BX (Given)
\[\therefore \frac{AX}{BX} = \frac{3}{\boxed{2}} \ \\ \frac{AX + BX}{BX} = \frac{3 + 2}{2} \quad \text{(by componendo)} \ \\ \frac{BA}{BX} = \frac{5}{2} \quad \dots \text{(I)} [6pt] \\ \text{Now } \triangle BCA \sim \triangle BYX ; (\boxed{\text{AA}} \text{ test of similarity}) [4pt] \\ \therefore \frac{BA}{BX} = \frac{AC}{XY} \quad \text{(corresponding sides of similar triangles)} [4pt] \\ \frac{5}{2} = \frac{AC}{9} \quad \text{from (I)} [4pt] \\ \therefore AC = \boxed{22.5}\]