Solve As shown in Figure 6, since $\triangle A B P$ and $\triangle A C Q$ have equal heights and bases, hence $S \triangle \triangle B P=S, \triangle C=$
$\triangle A B P$ and $\triangle A C Q$ have the same vertex angle $\triangle B A P$ $=\angle C A Q$,
$$\begin{array}{ll} X & B Q=B P+P Q, \quad C P=C Q+P Q, \\ \therefore \quad & B Q=C P . \\ \because \quad & \angle B A Q=\angle B A P+\angle P A Q, \\ & \angle C A P=\angle C A Q+\angle P A Q, \\ \therefore \quad & \angle B A Q=\angle C A P . \end{array}$$

From $\left\{\begin{array}{l}A B \cdot A P=1, \\ A C \cdot A Q \\ A B \cdot A Q=1 \\ \overline{A C \cdot A P}=1\end{array}\right.$
we get $\frac{A B^{2}}{A C^{2}}=1, A B=A C$,
which means $\triangle A B C$ is an isosceles triangle.