Problem 1451

1. Given Setup and Definitions:
- Points $A, B, C, D, E, F$ lie in that order on a semicircle centered at $O$.
- $AD = BE = CF$.
- $G$ is the intersection of $BE$ and $AD$.
- $H$ is the intersection of $BE$ and $CD$.

2. Objective:
- Prove that $\angle AOC = 2\angle GOH$.

3. Using the Power of a Point Theorem:
- The Power of a Point theorem states that for a point $P$ and a circle with center $O$ and radius $R$, if $P$ lies outside the circle and two lines through $P$ intersect the circle at points $A, B$ and $C, D$ respectively, then $PA \cdot PB = PC \cdot PD$.

4. Applying the Power of a Point Theorem:
- Consider the intersections $G$ and $H$ and their respective powers with respect to the semicircle.
- Since $G$ lies on both $AD$ and $BE$, we have:
$$GA \cdot GD = GB \cdot GE$$
- Similarly, since $H$ lies on both $BE$ and $CD$, we have:
$$HB \cdot HE = HC \cdot HD$$

5. Using the Given Lengths:
- Given $AD = BE = CF$, we can infer that the segments are equal in length.
- This implies that the points $A, D, B, E, C, F$ are symmetrically placed on the semicircle.

6. Angle Relationships:
- Since $A, B, C, D, E, F$ lie on a semicircle, the angles subtended by these points at the center $O$ are related.
- Let $\angle AOC = \theta$. We need to show that $\theta = 2\angle GOH$.

7. Central and Inscribed Angles:
- The central angle $\angle AOC$ is twice the inscribed angle subtended by the same arc $AC$.
- Therefore, if $\angle AOC = \theta$, then the inscribed angle $\angle AGC = \frac{\theta}{2}$.

8. Intersection Angles:
- Since $G$ and $H$ are intersections of the chords, the angles $\angle GOH$ can be related to the inscribed angles.
- Specifically, $\angle GOH$ is half of the central angle subtended by the arc between $G$ and $H$.

9. Conclusion:
- By the properties of the semicircle and the symmetry of the points, we can conclude that:
$$\angle AOC = 2\angle GOH$$

$\blacksquare$