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

Question

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

Answer 1

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?

Answer 2

Elegant Solution

Define point $E$ on the side $AD$ such that $|AE|=|BC|$. Now you have $ABE \equiv CBD$. Therefore $|BE| = |BD|.$ Now define the point $F$ as the mid point of $ED$ since $|BE| = |BD|$, we have $BF$ perpendicular to $AD$. A very nice observation here is $$ |AF| = \frac{|AD| + |AE|}{2} = \frac{|AD| + |BC|}{2} = = \frac{640 - 2(180)}{2} = 140 $$ Hence $$ \cos \alpha = \frac{|AF|}{180} = \frac{140}{180} $$