1.26. The area of an isosceles triangle is equal to $\frac{1}{3}$ of the area of the square constructed on the base of the given triangle. The lengths of the lateral sides of the triangle are shorter than the length of its base by 1 cm. Find the lengths of the sides and the height of the triangle, drawn to the base.
1.26. By the condition, $B C^{2}=$ $=3 \cdot \frac{1}{2} B C \cdot A H$ (Fig. 1.22), or $A H=$ $=\frac{2}{3} B C$. But $A H^{2}=A B^{2}-\left(\frac{1}{2} B C\right)^{2}$ and, therefore, $\frac{4}{9} B C^{2}=A B^{2}-\frac{1}{4} B C^{2}$, or $A B^{2}=\frac{25}{36} B C^{2}$, i.e., $A B=\frac{5}{6} B C$. Then we get $A B=\frac{5}{6}(A B+1)$, from which $A B=5$ (cm). Therefore, $B C=6 \mathrm{~cm}, A H=4 \mathrm{~cm}$.

Answer: 5 and 6 cm; 4 cm.