In the quadrilateral $A B C D$, there are right angles at vertices $B$ and $D$. The diagonal $A C$, which is one unit long, divides the angle at vertex $A$ into two acute angles $\alpha$ and $\beta$. What is the length of the other diagonal? - Let the reflection of vertex $B$ over the diagonal $A C$ be $B^{\prime}$. What is the length of the segment $B^{\prime} D$?
Let $\angle BAC = \alpha, \angle CAD = \beta$. Choose the labeling such that $\alpha \leq \beta$. Points $B$ and $D$ are on the Thales circle above the diameter $AC$, and due to reflection, $B'$ is also on the circle. Let the point opposite to $B$ and $B'$ in the circle $k$ be $B_1$ and $B_1'$, respectively, so $BB_1 = B'B_1' = AC = 1$. In the right triangle $BDB_1$, $\angle DB_1B = \angle DAB = \alpha + \beta$, or $= \angle DCB = 180^\circ - (\alpha + \beta)$, depending on whether $B_1$ is on the arc $BD$ containing $A$ or $C$, assuming the quadrilateral is convex. Thus, in any case, $BD = \sin(\alpha + \beta)$. (If $B_1$ coincides with $D$, then $BD = 1$, the diameter.)

![](https://cdn.mathpix.com/cropped/2024_05_02_11974f081f1425515200g-1.jpg?height=569&width=486&top_left_y=393&top_left_x=814)

Furthermore, in the right triangle $B'B_1'D$,

$$
\angle B'B_1'D = \angle B'AD = \angle DAC - \angle B'AC = \beta - \angle BAC = \beta - \alpha,
$$

and thus $B'D = \sin(\beta - \alpha)$, and if $B$ and $D$ are reflections of each other, then $B' = D$ and $B'D = 0$.

If $ABCD$ is a cyclic quadrilateral, or $AB$ and $AD$ are on the same side of the diagonal $AC$, then $\angle BAD = \beta - \alpha$, and similarly $BD = \sin(\beta - \alpha)$ and $B'D = \sin(\alpha + \beta)$.

Eszter Éltes (Budapest, I. István g., III. o. t.)