8,9 Find the perimeter of quadrilateral $A B C D$, in which $A B=C D=a, \angle B A D=\angle B C D=\alpha<90^{\circ}, B C \neq A D$.
Apply the Law of Cosines.

## Solution

Let $A D=x, B C=y$. By the Law of Cosines in triangles $A B D$ and $C B D$, we find that

$$
B D^{2}=a^{2}+x^{2}-2 a x \cdot \cos \alpha, B D^{2}=a^{2}+y^{2}-2 a y \cdot \cos \alpha
$$

therefore

$$
a^{2}+x^{2}-2 a x \cdot \cos \alpha=a^{2}+y^{2}-2 a y \cdot \cos \alpha
$$

Then

$$
x^{2}-y^{2}=(x-y) \cdot 2 a \cdot \cos \alpha
$$

since $x \neq y$, we have $x+y=2 a \cdot \cos \alpha$. Therefore,

$$
A B+C D+A D+B C=a+a+x+y=2 a+2 a \cdot \cos \alpha
$$

## Answer

$2 a(1+\cos \alpha)$