In $\triangle ABC$, $DE \parallel BC$. We are given that $\frac{AD}{DB} = \frac{3}{5}$ and $AC = 5.6$ cm.
By the Basic Proportionality Theorem (also known as Thales' Theorem), we have $$ \frac{AD}{DB} = \frac{AE}{EC} $$ Also, we have $\frac{AD}{AB} = \frac{AE}{AC}$. We are given $\frac{AD}{DB} = \frac{3}{5}$. Then $$ \frac{AD}{AD + DB} = \frac{AD}{AB} = \frac{3}{3+5} = \frac{3}{8} $$ Since $\frac{AD}{AB} = \frac{AE}{AC}$, we have $$ \frac{3}{8} = \frac{AE}{5.6} $$ So, $$ AE = \frac{3}{8} \times 5.6 = \frac{3}{8} \times \frac{56}{10} = \frac{3 \times 7}{10} = \frac{21}{10} = 2.1 $$ Therefore, $AE = 2.1$ cm.
In the adjoining figure, \( AP = 1 \, \text{cm}, \ BP = 2 \, \text{cm}, \ AQ = 1.5 \, \text{cm}, \ AC = 4.5 \, \text{cm} \) Prove that \( \triangle APQ \sim \triangle ABC \).
Hence, find the length of \( PQ \), if \( BC = 3.6 \, \text{cm} \).