Let $\triangle ABC$ be an isosceles.$(AB=AC)$.Let $D$ and $E$ be two points on the side $BC$ such that $D\in BE$,$E\in DC$ and $2\angle DAE = \angle BAC$.Prove that we can construct a triangle $XYZ$ such that $XY=BD$,$YZ=DE$ and $ZX=EC$.Find $\angle BAC + \angle YXZ$.
1. Given that $\triangle ABC$ is an isosceles triangle with $AB = AC$. Let $D$ and $E$ be points on $BC$ such that $D \in BE$, $E \in DC$, and $2\angle DAE = \angle BAC$.
2. We need to prove that we can construct a triangle $XYZ$ such that $XY = BD$, $YZ = DE$, and $ZX = EC$. Additionally, we need to find $\angle BAC + \angle YXZ$.

3. First, note that $\angle BAD + \angle CAE = \angle DAE$ because $D$ and $E$ lie on $BC$ and $2\angle DAE = \angle BAC$.

4. Consider folding $\triangle ABD$ along $AD$ and $\triangle AEC$ along $AE$. Let the folded triangles be $\triangle ADB'$ and $\triangle AEC'$, respectively.

5. By the given condition, $\angle BAD + \angle CAE = \angle DAE$. This implies that the folded triangles $\triangle ADB'$ and $\triangle AEC'$ will touch at a common point, say $B' = C'$.

6. Therefore, $\triangle DEB'$ can be created with $B' = C'$. This implies that $B'$ and $C'$ coincide, and thus $\triangle DEB'$ is a valid triangle.

7. Now, we need to find $\angle BAC + \angle YXZ$. Since $B' = C'$, we have:
   \[
   \angle BAC + \angle YXZ = \angle BAC + \angle DB'E
   \]
   Since $B'$ and $C'$ coincide, $\angle DB'E = \angle DB'A + \angle EB'A$.

8. Notice that $\angle DB'A = \angle ABC$ and $\angle EB'A = \angle ACB$ because $B'$ and $C'$ coincide with $B$ and $C$ respectively.

9. Therefore,
   \[
   \angle BAC + \angle YXZ = \angle BAC + \angle ABC + \angle ACB
   \]

10. Since $\triangle ABC$ is a triangle, the sum of its internal angles is $180^\circ$. Hence,
    \[
    \angle BAC + \angle ABC + \angle ACB = 180^\circ
    \]

11. Thus,
    \[
    \angle BAC + \angle YXZ = 180^\circ
    \]

The final answer is $\boxed{180^\circ}$.