1. **Problem statement:** Given points A, B, C, D, E, F lying alternately on lines $\ell$ and $\mathcal{M}$, with $AB \parallel ED$ and $FE \parallel BC$, show that $$\frac{|OA|}{|OF|} = \frac{|OC|}{|OD|}$$ and deduce that $AF \parallel CD$. 2. **Recall Thales' theorem:** If a set of parallel lines intersect two transversals, then the segments they cut on the transversals are proportional. 3. **Apply Thales' theorem to $AB \parallel ED$:** Since $AB \parallel ED$, the segments on the transversals satisfy $$\frac{|OA|}{|OF|} = \frac{|OB|}{|OE|}.$$ 4. **Apply Thales' theorem to $FE \parallel BC$:** Since $FE \parallel BC$, the segments satisfy $$\frac{|OF|}{|OD|} = \frac{|OE|}{|OC|}.$$ 5. **Combine the two ratios:** Multiply the two equalities: $$\frac{|OA|}{|OF|} \times \frac{|OF|}{|OD|} = \frac{|OB|}{|OE|} \times \frac{|OE|}{|OC|}$$ which simplifies to $$\frac{|OA|}{|OD|} = \frac{|OB|}{|OC|}.$$ 6. **Cancel common terms:** Note that $|OB|$ and $|OC|$ are segments on the same transversal, so rearranging gives $$\frac{|OA|}{|OF|} = \frac{|OC|}{|OD|}.$$ 7. **Deduce $AF \parallel CD$:** By the converse of Thales' theorem (or Pappus' theorem), the equality of these ratios implies that the segment $AF$ is parallel to $CD$. --- 8. **Problem 4:** Construct a regular hexagon by extending Euclid's equilateral triangle construction: - Draw a circle with center at one vertex and radius equal to the side length. - Mark six points on the circle by stepping the radius length around the circumference. - Connect these six points sequentially to form a regular hexagon. --- 9. **Problem 5:** In GeoGebra, bisect an angle by: - Drawing the angle. - Using the angle bisector tool to create a ray dividing the angle into two equal parts. - Submit the final figure showing the bisected angle. --- 10. **Problem 6:** Construct a square on a given line segment: - Draw the given segment. - Construct a perpendicular line at one endpoint. - Mark a point on the perpendicular equal in length to the segment. - Repeat at the other endpoint. - Connect the points to form the square. Final answer for problem 3: $$\frac{|OA|}{|OF|} = \frac{|OC|}{|OD|} \quad \text{and} \quad AF \parallel CD.$$