12.197. 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 $Z$. Find the ratio of the area of triangle $K Z M$ to the area of triangle $A B C$, if $\angle A=\alpha$ and $\angle C=\beta$.
Solution.

$\angle A B C=180^{\circ}-(\alpha+\beta)($ Fig. 12.64)

$D A M B$ and $\triangle M K B, \triangle C M B$ and $\triangle M Z B$ are right triangles with common acute angles at vertex $B$.

Thus, $\angle K M B=\angle A=\alpha, \angle Z M B=\angle C=\beta, \angle K M Z=\alpha+\beta$. Let $B M=1$, then from $\triangle A M B\left(\angle A M B=90^{\circ}\right)$ :

$A B=\frac{B M}{\sin \angle A}=\frac{1}{\sin \alpha} ;{ }_{\text {in }} \triangle C M B\left(\angle C M B=90^{\circ}\right): B C=\frac{B M}{\sin \angle C}=\frac{1}{\sin \beta}$;

From $\triangle B K M\left(\angle B K M=90^{\circ}\right): M K=B M \cos \angle K M B=\cos \alpha$;

In $\triangle B Z M\left(\angle B Z M=90^{\circ}\right): M Z=B M \cos \angle Z M B=\cos \beta$. Therefore,

$\frac{S_{\triangle K M Z}}{S_{\triangle A B C}}=\frac{\frac{1}{2} M K \cdot M Z \sin \angle K M Z}{\frac{1}{2} B A \cdot B C \sin \angle A B C}=\frac{\cos \alpha \cos \beta \sin (\alpha+\beta)}{\frac{1}{\sin \alpha} \cdot \frac{1}{\sin \beta} \sin \left(180^{\circ}-(\alpha+\beta)\right)}=$

$=\sin \alpha \cos \alpha \sin \beta \cos \beta=\frac{1}{4} \sin 2 \alpha \sin 2 \beta$.

Answer: $\frac{1}{4} \sin 2 \alpha \sin 2 \beta$.