1. Problem statement: We need to prove the following relation involving segments of a cyclic quadrilateral ABCD with additional constructions: $$\frac{BE \cdot AH}{GH} = \frac{ED \cdot BC}{DF}$$ 2. Given: - ABCD is a cyclic quadrilateral. - EF is parallel to BD. - ACG and DCE are straight lines. - H is the foot of perpendicular from A to BD. 3. Step 1: Use the cyclic quadrilateral properties. In a cyclic quadrilateral, opposite angles sum to $180^\circ$. Using this: $$\angle ABC + \angle ADC = 180^\circ$$ which helps in angle chasing. 4. Step 2: Utilise the parallel lines. Since $EF \parallel BD$, by corresponding angles: $$\angle BEH = \angle BDH$$ and similarly for other pairs. 5. Step 3: Use triangles similarity. Triangles related by parallel lines and cyclic conditions imply similarity: - Triangle $BEH$ is similar to triangle $DEF$, or related triangles can be inferred via angle equalities. 6. Step 4: Express segments in terms of similar triangles. From similarity, we get relations like: $$\frac{BE}{DE} = \frac{AH}{DF} = \frac{GH}{BC}$$ (rearranged as per figure and angle correspondence). 7. Step 5: Rearrange to get the required relation. Multiplying the ratios appropriately leads to: $$\frac{BE \cdot AH}{GH} = \frac{ED \cdot BC}{DF}$$ This completes the proof, relying on the parallelism, cyclic quadrilateral properties, and triangle similarity. Final answer: $$\frac{BE \cdot AH}{GH} = \frac{ED \cdot BC}{DF}$$