3. Let $H$ be the orthocenter of isosceles $\triangle A B C$. Keeping the base $B C$ unchanged, if the distance from vertex $A$ to the base $B C$ decreases, what happens to the product $S_{\triangle A B C} \cdot S_{\triangle H B C}$? Does it decrease, increase, or remain unchanged?
3. Unchanged. Divide $\angle A$ into three cases: acute, right, and obtuse angles, and discuss each respectively. Or set the circumradius of $\triangle A B C$ to 1. By property 5 of the orthocenter, we have $C H=B H=2 \cos B \cdot S_{\triangle A B C}=\frac{1}{2} C H \cdot B H \cdot \sin \angle B H C=\frac{1}{2} B H \cdot C H \cdot \sin \left(180^{\circ}-\angle A\right)=2 \sin A \cdot \cos ^{2} B$. Also, $S_{\triangle A B C}=\frac{1}{2} A B \cdot A C \cdot \sin A=2 \sin A \cdot \sin ^{2} B$, so $S_{\triangle A B C} \cdot S_{\triangle H B C}=\sin ^{2} A \cdot \sin ^{2} 2 B$, and since $\sin 2 B=\sin A$, we have $S_{\triangle A B C} \cdot S_{\triangle H B C}=\frac{1}{16} B C^{4}$ (unchanged).