In the convex quadrilateral $A B C D$, the angles at vertices $A$ and $B$ are equal, the angle at $C$ is a right angle. The side $A D$ is perpendicular to the diagonal $B D$. The length of side $B C$ is equal to the length of side $C D$. How many times the common length is the length of side $A D$?
Solution. The quadrilateral can be obtained by joining the right-angled triangles $ABD$ and $BCD$ along their side $BD$. According to the condition, $BC = CD = a$, so $BCD$ is an isosceles right-angled triangle, $\angle CBD = \angle CDB = 45^{\circ}$, and $BD = a \sqrt{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_23d92ac1d8695697f178g-1.jpg?height=523&width=420&top_left_y=302&top_left_x=842)

Extend the sides $AD$ and $BC$, and let their intersection point be $M$. Since $\angle DAB = \angle ABC = \varphi$, the triangle $ABM$ is also isosceles.

In the quadrilateral $ABCD$,

$$
2 \varphi + 90^{\circ} + (45^{\circ} + 90^{\circ}) = 360^{\circ}
$$

From this, $2 \varphi = 135^{\circ}$ and $\angle AMB = 45^{\circ}$. Then $DCM$ is also an isosceles right-angled triangle and $DM = a \sqrt{2}$. We seek the ratio $\frac{DC}{AD} = \frac{a}{x}$. The leg of the isosceles triangle $ABM$ is $x + a \sqrt{2} = 2a$. Rearranging the equation, we get $a(2 - \sqrt{2}) = x$, so

$$
\frac{a}{x} = \frac{1}{2 - \sqrt{2}} \approx 1.7071
$$