Problem 37. The base of an isosceles triangle relates to its lateral side as 4:3 (Fig. 28). In what ratio does the height, drawn to the lateral side, divide it?
Solution. According to formula II:

$$
\begin{aligned}
& 4 x=\sqrt{2 \cdot 3 x \cdot m} \text {, or } 16 x^{2}=6 x m, \\
& \text { hence } m=\frac{8}{3} x . \text { Then } \\
& \quad n=3 x-\frac{8}{3} x=\frac{1}{3} x, \text { therefore, } n: m=1: 8 .
\end{aligned}
$$

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