# How to solve for x, when you know quadrilateral's sides and angles created by the diagonal?

This problem was asked from 8th and 9th graders in a contest.

You have an quadrilateral $$ABCD$$ with $$AB=4x-y$$, $$BC=3x-4$$, $$CD=y$$, and $$DA=5x-2$$. You draw a line $$BD$$ which makes a right triangle with $$\angle CBD=36°$$ and in the triangle $$ABD$$ $$\angle DBA=126°$$. Solve for $$x$$.

I was able to solve this with a calculator but this problem is supposed to be solved without one. I tried to find an answer to this problem online but I couldn't find one, probably because my math vocab in English isn't that great.

Nice question!

Reflect the upper left triangle in the diagonal and rotate the lower triangle about the mid-point of the diagonal. (Delete the original triangles.)

You should now be able to see a Pythagorean triangle with sides $$3x-4,4x,5x-2$$.

Then $$x=3$$ gives the solution and the Pythagorean triangle is a $$(5,12,13)$$ triangle.

Considering the right triangle BCD, one can write :

$$y=(3x-4)\tan(36°)\tag{1}$$

Let us apply to the other triangle ABD the law of cosines:

$$AD^2=AB^2+BD^2-2 AB.BD \cos(126°)$$

$$=AB^2+\underbrace{(BC^2+CD^2)}_{\text{Pythagoras}}-2 AB.\dfrac{BC}{\cos(36°)} (-\sin(36°))$$

$$(5x-2)^2=(4x-y)^2+y^2+(3x-4)^2+2 (4x-y).\underbrace{(3x-4)\tan(36°)}_{= \ y \ \ \text{using} \ (1)} \ \ \ \ \ (2)$$

Expanding (2), which is an identity, one obtains $$4x-12=0$$, imposing value

$$x=3.$$