In a ∆ABC, D and E are points on AB and AC respectively such that DE || BC.

It is given that $A D=2.4 \mathrm{~cm}, A E=3.2 \mathrm{~cm}, D E=2 \mathrm{~cm}$ and $B C=5 \mathrm{~cm}$.

We have to find BD and CE.

Since DE∥BC, AB is transversal, then

∠ADE = ∠ABC   (corresponding angles)

Since DE∥BC, AC is a transversal, then

∠AED = ∠ACB   (corresponding angles)

In ∆ADE and ∆ABC,

∠ADE = ∠ABC  (proved above)

∠AED = ∠ACB  (proved above)

so, ∆ADE ∼ ∆ABC (Angle Angle similarity)

Since, the corresponding sides of similar triangles are proportional, then

$\mathrm{ADAB}=\mathrm{AEAC}=\mathrm{DEBC} \Rightarrow \mathrm{ADAB}=\mathrm{DEBC} \Rightarrow 2.42 .4+\mathrm{DB}=25 \Rightarrow 2.4+\mathrm{DB}=6 \Rightarrow \mathrm{DB}=6-2.4 \Rightarrow \mathrm{DB}=3.6 \mathrm{~cm}$ Similarly, $\quad \mathrm{AEAC}=\mathrm{DEBC} \Rightarrow 3.23 .2+\mathrm{EC}=25 \Rightarrow 3.2+\mathrm{EC}=8 \Rightarrow \mathrm{EC}=8-3.2 \Rightarrow \mathrm{EC}=4.8 \mathrm{~cm}$

Hence, BD = 3.6 cm  and  CE = 4.8 cm.