<s>[INST] In convex quadrilateral $ABCD,$ $\angle A \cong \angle C,$ $AB=CD=180,$ and $AD \ne BC.$ The perimeter of $ABCD$ is $640.$ Find $\cos A.$ In convex quadrilateral $ABCD,$ $\angle A \cong \angle C,$ $AB=CD=180,$ and $AD \ne BC.$ The perimeter of $ABCD$ is $640.$ Find $\cos A.$

I have no idea on where to start on this problem
 [/INST] Hint: Let $AD = x$ and $BC = y$. Then $x+y+180+180 = 640$ so $x+y = 280$. By the Law of Cosines, we have
$$ BD^2 = AD^2+AB^2-2AD\cdot AB\cos\alpha = x^2+180^2-360x\cos\alpha $$
and
$$ BD^2 = BC^2+CD^2-2BC\cdot CD\cos\alpha = y^2+180^2-360y\cos\alpha. $$
Therefore
$$0 = x^2-y^2 - 360(x-y)\cos\alpha. $$
Can you take it from here?
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