Let $\omega$ be the circle of center $A$ and radius $AB$. Let $A'$ be a point on the circle $\omega$, lying on the minor arc $\widehat{BC}$, such that $\angle BAD = \angle A'AD$. Since $2\angle DAE = \angle A$, it is easy to see that $\angle CAE = \angle A'AE$.

We deduce that the triangles $BAD$ and $A'AD$ are congruent by SAS postulate, and thus $BD = A'D$. Similarly, $EC = A'E$, and thus, the triangle $A'DE$ has its sides of lengths $BD$, $DE$ and $EC$ respectively, and we can choose $X = A'$, $Y = D$ and $Z = E$.

Moreover,
$$\begin{align*} \angle BAC + \angle YXZ &= \angle BAC + \angle DA'E \\ &= \angle BAC + \angle DA'A + \angle AA'E \\ &= \angle BAC + \angle DBA + \angle ACE = 180^{\circ}. \end{align*}$$

Let $M$ be the midpoint of $BC$. Obviously $D \in BM$ and $E \in MC$. We make now the following notations $\angle A = 2t$, $BC = 2a$, $BD = x$, $EC = z$ and $DE = y$. It is clear that $DE = 2a - x - z$.

On the other hand, $AM = a \cdot \cot t$. But $t = \angle DAM + \angle MAE$, and thus we get:
$$\cot(t) = \frac{\cot \angle DAM \cdot \cot \angle MAE - 1}{\cot \angle DAM + \cot \angle MAE} \quad (1)$$
Denoting $x + z = s$ and $x \cdot z = p$, using the relation (1), we get:
$$a^2(1 + \cot^2 t) = a \cdot s(1 + \cot^2 t) - p.$$
Now, from the above relation, we get:
$$s = a + \frac{p}{a} \sin^2 t.$$
Finally, we obtain:
$$y^2 = (2a - s)^2 = x^2 + 2x \cdot z \cos(2t) + z^2.$$
Now, on the two sides of an angle of the vertex $X$ and measure $\pi - A$, we choose the points $Y$ and $Z$ such that $XY = x$ and $XZ = z$. From the cosine rule we have:
$$YZ^2 = x^2 + 2x \cdot z \cos(2t) + z^2 = y^2$$
and thus the existence of the triangle $XYZ$ is proved. Moreover, we have $\angle BAC + \angle YXZ = 180^{\circ}$.