In ∆ABC, P and Q are points on sides AB and AC respectively

In triangle $A B C, P$ and $Q$ are points on sides $A B$ and $A C$ respectively such that $P Q \| B C$.

In $\triangle A P Q$ and $\triangle A B C$,

$\angle A P Q=\angle B$  (Corresponding angles)

$\angle P A Q=\angle B A C \quad$ (Common)

So, $\triangle \mathrm{APQ}-\Delta \mathrm{ABC}$ (AASimilarity)

$\frac{A P}{A B}=\frac{P Q}{B C}$

Substituting value $A P=3 \mathrm{~cm}, A B=10 \mathrm{~cm}$ and $P Q=3 \mathrm{~cm}$, we get

$\frac{4}{10}=\frac{3}{B C}$

By cross multiplication we get

$4 \times B C=3 \times 10$

$B C=\frac{3 \times 10}{4}$

$B C=\frac{30}{4}$

$B C=7.5 \mathrm{~cm}$

Hence, the value of $B C$ is $7.5 \mathrm{~cm}$.