The unit-sided square $ABCD$ has its midpoint $F$ on side $AB$. A circle passing through $C$ and centered at $F$ intersects the ray $AB$ at $G$. The circle with diameter $AG$ intersects the segment $BC$ at $H$. With what precision does the segment $BH$ approximate the length of the semicircular arc with diameter $BF$?
Based on the Pythagorean theorem, $F G=F C=\sqrt{5} / 2$, therefore $B G=(\sqrt{5}-1) / 2$, and finally from the right triangle $A G H$

$$
B H=\sqrt{A B \cdot B G}=\sqrt{\frac{\sqrt{5}-1}{2}}=0.78615 \ldots
$$

On the other hand, the length of the questioned semicircular arc is $\pi / 4=0.78539 \ldots$, which is approximated from above by the segment $B H$, their ratio is $1.000 \quad 959<1.001$. The difference is less than $1\%$.

![](https://cdn.mathpix.com/cropped/2024_05_02_9e84484f1d32b3ec8017g-1.jpg?height=456&width=647&top_left_y=485&top_left_x=739)

Remark. The point $H$ is also intersected from $B C$ by the unit circle centered at $G$. Furthermore, the line $H-\mathrm{n}$ also passes through the point $K$ on the line $A K$, which is intersected from the ray $D C$ by the circle centered at $A$ passing through $G$. [See: Tusnády Gábor: The Golden Section, K. M. L. 57 (1978) p. 72.]