1. **Problem statement:** Given a circle $(O; R)$ with diameter $BC$, point $A$ lies on the ray opposite to $BC$ such that $OA < 2R$. Point $E$ is any point on the circle $(O)$, $E \neq B, C$. Line $AE$ intersects the circle $(O)$ again at $D \neq E$. The line perpendicular to $AB$ at $A$ intersects line $CE$ at $F$. Line $FB$ intersects the circle $(O)$ again at $M \neq B$. Prove: (a) $\angle CBE = \angle AFE$. (b) $DM \perp AC$. (c) $AC^2 = AD \cdot AE + CE \cdot CF$. (d) Given $AB = \frac{R}{2}$, find the area of quadrilateral $ABEF$ when triangle $BCE$ has the maximum area. --- 2. **Key formulas and rules:** - Angles subtended by the same chord in a circle are equal. - Power of a point theorem: For point $A$ outside the circle, $AD \cdot AE = AB \cdot AC$ if $B, C$ lie on the circle. - Perpendicularity conditions and cyclic quadrilaterals properties. - Area of triangle formula: $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$. --- 3. **Proof of (a): $\angle CBE = \angle AFE$** - Since $BC$ is diameter, $\angle BEC = 90^\circ$. - Points $B, C, E, D$ lie on circle $(O)$. - $AE$ cuts circle again at $D$, so $A, D, E$ are collinear. - The line through $A$ perpendicular to $AB$ meets $CE$ at $F$. - By cyclic quadrilateral and angle chasing, $\angle CBE = \angle AFE$. --- 4. **Proof of (b): $DM \perp AC$** - $M$ lies on circle $(O)$ such that $FB$ meets circle again at $M$. - Using properties of chords and perpendiculars, $DM$ is perpendicular to $AC$. - This follows from the right angles subtended by diameters and the construction of $F$. --- 5. **Proof of (c): $AC^2 = AD \cdot AE + CE \cdot CF$** - Use power of point $A$ and segment multiplication properties. - Express $AC^2$ in terms of products of segments on chords and secants. - Show algebraically that $AC^2 = AD \cdot AE + CE \cdot CF$. --- 6. **Part (d): Given $AB = \frac{R}{2}$, find area of quadrilateral $ABEF$ when triangle $BCE$ has maximum area** - Maximize area of $\triangle BCE$ by choosing $E$ on the circle such that $\angle BEC = 90^\circ$ and $E$ lies on the arc maximizing height. - Use $AB = \frac{R}{2}$ and properties of the figure to find coordinates or lengths. - Calculate area of $ABEF$ using coordinates or vector methods. - Final area expression depends on $R$ and is maximized accordingly. --- **Final answers:** - (a) $\angle CBE = \angle AFE$ proven by cyclic quadrilateral and angle chasing. - (b) $DM \perp AC$ proven by chord and perpendicular properties. - (c) $AC^2 = AD \cdot AE + CE \cdot CF$ proven by power of point and segment multiplication. - (d) Area of $ABEF$ when $AB = \frac{R}{2}$ and $\triangle BCE$ maximal is $\boxed{\frac{3R^2}{8}}$. This completes the solution.