In \(\triangle ABC\), if \(DE \parallel BC\), \(AE/CE = 3/5\) and \(AB = 5.6\) cm, then \(AD =\):
Show Hint
In problems involving parallel lines and triangles, use the Basic Proportionality Theorem (Thales' theorem) to relate the corresponding sides of the triangle.
In a triangle, if a line parallel to one side divides the other two sides in a given ratio, then by the Basic Proportionality Theorem (also known as Thales’ theorem), we have:
\[
\frac{AE}{CE} = \frac{AB}{BC}
\]
We are given that \(\frac{AE}{CE} = \frac{3}{5}\), and \(AB = 5.6\) cm. Therefore:
\[
\frac{AB}{BC} = \frac{3}{5}
\]
Let \(BC = x\). Then:
\[
\frac{5.6}{x} = \frac{3}{5}
\]
Cross-multiplying:
\[
5.6 \times 5 = 3 \times x \quad \Rightarrow \quad x = \frac{28}{3} \approx 9.33
\]
Now, using the relation \(\frac{AD}{AB} = \frac{AE}{AB}\):
\[
AD = \frac{AE}{AB} \times AB
\]
Thus, the correct answer is option (1), \(AD = 2.8\) cm.
