## Solution 2.

Let us consider a parallelogram $A B C D$, with $A B=a, B C=b, A C=x, B D=y$, $\widehat{B O C}=\theta$, and let us produce the line $A D$ towards $D$ and consider $M \in(A D$ so that $A D=D M$. Then $B C M D$ is a parallelogram, so $C M=B D=y$.

Observe also that $(A B C D)=2(A C D)=(A C M)$ which is written equivalently as

$$C B \cdot C D \cdot \sin C=\frac{A C \cdot C M \cdot \sin \theta}{2} \text { i.e. } a b \sin C=\frac{x y \sin \theta}{2}$$

Because of the given relation $a b=\frac{x y}{2}$ the last relation becomes $\sin C=\sin \theta$, i.e.

$$\theta=\widehat{C} \text { or } \theta=180^{\circ}-\widehat{C}=\widehat{B}$$

If $\theta=\widehat{C}$, then the triangles $A C M$ and $B C D$ are similar because their angles at $C$ are equal, as well as their angles at $B, M$ (remember $B C M D$ is a parallelogram).

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Then

$$\frac{b}{y}=\frac{a}{x}=\frac{y}{2 b} \Rightarrow\left(b=\frac{y}{2}, a=\frac{x}{2}\right)$$

If $\theta=\widehat{B}$, then similarly we prove that the triangles $A C M$ and $A C D$ are similar, which then implies

$$\frac{a}{y}=\frac{b}{x}=\frac{x}{2 b} \Rightarrow\left(a=\frac{y}{2}, b=\frac{x}{2}\right)$$