Problem 6.1. Given a quadrilateral $A B C D$, in which $A D \| B C$. The bisector of angle $A$ intersects side $C D$ at point $X$, and the extension of side $B C$ beyond point $C$ - at point $Y$. It turns out that $\angle A X C=90^{\circ}$. Find the length of segment $A B$, if it is known that $A D=16$ and $C Y=13$.

![](https://cdn.mathpix.com/cropped/2024_05_06_2a01f5e31e9fc3823579g-10.jpg?height=499&width=809&top_left_y=1139&top_left_x=322)
Answer: 14.5.

Solution. $\quad$ Let $AB=a$. Note that $\angle BYA=\angle YAD=\angle YAB$. We get that triangle $ABY$ is isosceles $(AB=BY)$, and $BC=BY-CY=AB-CY=a-13$ (Fig. 1).

![](https://cdn.mathpix.com/cropped/2024_05_06_2a01f5e31e9fc3823579g-11.jpg?height=610&width=830&top_left_y=143&top_left_x=380)

Fig. 1: to the solution of problem 6.1

Extend rays $AB$ and $DC$ until they intersect at point $Z$. In triangle $ADZ$, the bisector $AX$ coincides with the altitude, so it is isosceles $(AZ=AD)$. Additionally, $\angle BZC=\angle ADC=\angle BCZ$, so triangle $BCZ$ is also isosceles $(BZ=BC)$.

Segment $AZ$, on the one hand, equals $AD$, which is 16. On the other hand, $AZ=AB+BZ=AB+BC=a+(a-13)=2a-13$. Solving the resulting equation:

$$
\begin{gathered}
16=2a-13 \\
2a=29 \\
a=14.5
\end{gathered}
$$