1. **Problem statement:** Given an acute triangle $\Delta ABC$ with $AB < AC$ inscribed in circle $(O; R)$. Altitudes $BK$ and $CH$ are drawn. The bisector of angle $\widehat{BAC}$ intersects $BC$ at $D$ and the minor arc $BC$ at $E$. Prove: (a) Quadrilateral $BHKC$ is cyclic and $OE \perp BC$. (b) $AD \cdot AE = AB \cdot AC$ and $AD^2 = AB \cdot AC - DB \cdot DC$. (c) Given $R=4$ cm, $BC=7.2$ cm, find the area of the lune bounded by chord $BC$ and the area of the ring bounded by circles $(A; AD)$ and $(A; AM)$, where $M$ is the reflection of $D$ about $E$. --- 2. **Part (a) Proof:** - To prove $BHKC$ is cyclic, note that $BK$ and $CH$ are altitudes, so $K$ and $H$ lie on the feet of altitudes from $B$ and $C$. - Since $\angle BKH = 90^\circ$ and $\angle BCH = 90^\circ$, quadrilateral $BHKC$ has opposite angles summing to $180^\circ$, so it is cyclic. - For $OE \perp BC$, $E$ lies on the bisector of $\widehat{BAC}$ and on the minor arc $BC$. - The radius $OE$ is perpendicular to chord $BC$ at $E$ by the property of circle radius perpendicular to chord. --- 3. **Part (b) Proof:** - By the Angle Bisector Theorem, $\frac{BD}{DC} = \frac{AB}{AC}$. - Using power of point $A$ with respect to circle $(O; R)$, $AD \cdot AE = AB \cdot AC$. - Also, by segment relations in triangle and chord properties, $$AD^2 = AB \cdot AC - DB \cdot DC$$ - This follows from the power of point and chord segment multiplication. --- 4. **Part (c) Calculations:** - Given $R=4$ cm, $BC=7.2$ cm. - Area of segment (lune) bounded by chord $BC$ and minor arc $BC$: - Compute central angle $\theta$ subtending chord $BC$: $$\cos \frac{\theta}{2} = \frac{BC}{2R} = \frac{7.2}{8} = 0.9$$ $$\Rightarrow \frac{\theta}{2} = \arccos(0.9) \approx 0.4510 \text{ rad}$$ $$\theta \approx 0.9020 \text{ rad}$$ - Area of sector $BOC$: $$S_{sector} = \frac{\theta}{2\pi} \pi R^2 = \frac{\theta}{2} R^2 = 0.4510 \times 16 = 7.216 \text{ cm}^2$$ - Area of triangle $BOC$: $$S_{triangle} = \frac{1}{2} R^2 \sin \theta = 0.5 \times 16 \times \sin(0.9020) \approx 8 \times 0.7833 = 6.2664 \text{ cm}^2$$ - Area of segment (lune): $$S_{lune} = S_{sector} - S_{triangle} = 7.216 - 6.2664 = 0.9496 \approx 0.9 \text{ cm}^2$$ - For the ring area between circles $(A; AD)$ and $(A; AM)$: - Since $M$ is reflection of $D$ about $E$, $AM = 2AE - AD$. - Using $AD \cdot AE = AB \cdot AC$ and $AB < AC$, approximate $AD$ and $AE$ from given data or use geometric relations. - Without explicit lengths $AB, AC$, assume $AD$ and $AE$ such that $AD < AE$. - The ring area is: $$S_{ring} = \pi (AM^2 - AD^2)$$ - Using $AM = 2AE - AD$, expand: $$AM^2 - AD^2 = (2AE - AD)^2 - AD^2 = 4AE^2 - 4AE \cdot AD + AD^2 - AD^2 = 4AE^2 - 4AE \cdot AD$$ - Substitute $AD \cdot AE = AB \cdot AC$: $$S_{ring} = \pi (4AE^2 - 4AB \cdot AC)$$ - Without $AE$ or $AB, AC$ values, numerical answer cannot be finalized. --- **Final answers:** - (a) $BHKC$ is cyclic and $OE \perp BC$. - (b) $AD \cdot AE = AB \cdot AC$ and $AD^2 = AB \cdot AC - DB \cdot DC$. - (c) Area of lune $\approx 0.9$ cm$^2$. Ring area depends on $AD, AE$ values not fully given.