11.151. The base of an oblique parallelepiped is a rhombus $ABCD$ with a side equal to $a$ and an acute angle of $60^{\circ}$. The edge $AA_{1}$ is also equal to $a$ and forms angles of $45^{\circ}$ with the edges $AB$ and $AD$. Determine the volume of the parallelepiped.
Solution.

Let $A_{1} O$ be the height of the parallelepiped $A B C D A_{1} B_{1} C_{1} D_{1}$ mentioned in the problem (Fig. 11.47). Since $\angle A_{1} A B = \angle A_{1} A D$, point $O$ lies on the bisector of $\angle B A D$ - the ray $A C$. $A A_{1}$ is an oblique to the base plane of the prism, $A O$ is its projection on this plane, and $A B$ is a line lying in this plane.

Draw $A_{1} K \perp A B$ and connect $K$ with $O$. In $\triangle A_{1} O A: \cos \angle A_{1} A O = \frac{O A}{A_{1} A}$; in $\triangle A_{1} K A: \cos \angle A_{1} A K = \cos \angle A_{1} A B = \frac{A K}{A_{1} A}$; in $\triangle O K A: \cos \angle O A K = \cos \angle O A B = \frac{A K}{O A}$. But $\frac{O A}{A_{1} A} \cdot \frac{A K}{O A} = \frac{A K}{A_{1} A}$, i.e., $\cos \angle A_{1} A O \cdot \cos \angle O A K = \cos \angle A_{1} A K$, from which $\cos \angle A_{1} A O = \frac{\cos \angle A_{1} A K}{\cos \angle O A K}$.

Since $\angle B A D = 60^{\circ}$, then $\angle O A K = 30^{\circ}$ and, therefore, $\cos \angle A_{1} A O = \frac{\cos 45^{\circ}}{\cos 30^{\circ}} = \frac{\sqrt{2}}{2} : \frac{\sqrt{3}}{2} = \frac{\sqrt{2}}{\sqrt{3}}$; $\sin \angle A_{1} A O = \sqrt{1 - \cos^2 \angle A_{1} A O} = \sqrt{1 - \frac{2}{3}} = \frac{1}{\sqrt{3}}$. Then $A_{1} O = A A_{1} \sin \angle A_{1} A O = \frac{a}{\sqrt{3}}$, from which we find the desired volume:

$V = S_{A B C D} \cdot A_{1} O = a^{2} \sin 60^{\circ} \cdot \frac{a}{\sqrt{3}} = \frac{a^{2} \sqrt{3}}{2} \cdot \frac{a}{\sqrt{3}} = \frac{a^{3}}{2}$.

Answer: $\frac{a^{3}}{2}$.