In the acute triangle $ABC$, $\angle BAC = 60^\circ$. The circle constructed on the side $BC$ as a diameter intersects the side $AB$ at $D$ and the side $AC$ at $E$. Determine the ratio of the area of the quadrilateral $BDEC$ to the area of triangle $ABC$.

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Since $D$ and $E$ lie on the Thales circle of segment $BC$, $\angle CDB = \angle CEB = 90^\circ$, which means $BE$ and $CD$ are altitudes of triangle $ABC$. We denote the orthocenter of the triangle by $M$. Since $\angle MEA = \angle MDA = 90^\circ$ and $\angle EAD = 60^\circ$, it follows that $\angle EMD = 360^\circ - (2 \cdot 90^\circ + 60^\circ) = 120^\circ$. Therefore, the angle between the diagonals of quadrilateral $BCED$ is $60^\circ$.

Triangle $ADC$ is a right triangle with one of its acute angles being $60^\circ$. If $AC = b$, then $CD = \frac{\sqrt{3}}{2} \cdot b$. Similarly, if $AB = c$, then $BE = \frac{\sqrt{3}}{2} \cdot c$. The ratio of the areas of the quadrilateral and the triangle is therefore:

$$
\frac{T_{BCED}}{T_{ABC}} = \frac{\frac{EB \cdot CD \cdot \sin 60^\circ}{2}}{\frac{AC \cdot AB \cdot \sin 60^\circ}{2}} = \frac{\frac{\sqrt{3}}{2} \cdot c \cdot \frac{\sqrt{3}}{2} \cdot b}{b \cdot c} = \frac{3}{4}
$$

Mile Veronika (Fazekas M. Főv. Gyak. Gimn., II. o.t.)

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