3.64. In triangle $ABC$, the altitude $BM$ is drawn, and a circle is constructed on it as a diameter, intersecting side $AB$ at point $K$ and side $BC$ at point $L$. Find the ratio of the area of triangle $KLM$ to the area of triangle $ABC$, if $\angle A=\alpha$ and $\angle C=\beta$.
3.64. By the condition, $B M \perp A C, B M$ is the diameter of the circle, $\angle B A C=\alpha, \angle B C A=\beta, K$ and $L$ are the points of intersection of the circle with the sides $A B$ and $B C$ (Fig. 3.66); it is required to find $S_{\triangle K L M}: S_{\triangle A B C}$. Let $B M=h$; then $A M=h \operatorname{ctg} \alpha, M C=h \operatorname{ctg} \beta$. We have

$$
\begin{gathered}
S_{\triangle A B C}=S_{\triangle A M B}+S_{\triangle B M C}=\frac{1}{2} A M \cdot B M+\frac{1}{2} M C \cdot B M= \\
=\frac{1}{2} B M(A M+M C)=\frac{1}{2} h(h \operatorname{ctg} \alpha+h \operatorname{ctg} \beta)= \\
=\frac{1}{2} h^{2}(\operatorname{ctg} \alpha+\operatorname{ctg} \beta)=\frac{1}{2} h^{2} \frac{\sin (\alpha+\beta)}{\sin \alpha \sin \beta}
\end{gathered}
$$

Further, $\angle A B M=90^{\circ}-\alpha, \angle M B C=90^{\circ}-\beta$ and since $\angle B K M=$ $=\angle B L M=90^{\circ}$ (as inscribed angles subtending the diameter $B M$), then $\angle B M K=\alpha, \angle B M L=\beta$. Then from $\triangle B K M$ and $\triangle B L M$ we find $K M=$ $=h \cos \alpha, M L=h \cos \beta$. Therefore,

$$
S_{\Delta K L M}=\frac{1}{2} K M \cdot M L \sin (\alpha+\beta)=\frac{1}{2} h^{2} \cos \alpha \cos \beta \sin (\alpha+\beta)
$$

Finally, we get

$$
\frac{S_{\triangle K L M}}{S_{\triangle A B C}}=\frac{0.5 h^{2} \cos \alpha \cos \beta \sin (\alpha+\beta) \sin \alpha \sin \beta}{0.5 h^{2} \sin (\alpha+\beta)}=\frac{1}{4} \sin 2 \alpha \sin 2 \beta
$$