In a rectangle ABCD there is a triangle ABP, with its vertex P on the CD side of the rectangle. The known measures of the sides are: $$AB = a$$ $$AD = (2\ -\sqrt3)a$$
I need to find the DAP triangle, knowing that it is valid the relation: $$AP^2+AD^2 = BP^2$$
I'm stuck with this and I'd really appreciate some suggestions or a solution.
Since $ABCD$ is a rectangle, $AD = BC$.
So, $\ AP^2+AD^2 = BP^2 = BC^2+PC^2 \Rightarrow AP = PC = x$.
By the Pythagorean Theorem for triangle $ADP$, we have that: $$ AD^2+DP^2 = AP^2 \Rightarrow (2-\sqrt{3})^2a^2+(x-a)^2 = x^2$$ Simplifying this equation gives: $$x = \frac{(2-\sqrt{3})^2+1}{2}a = (4-2\sqrt{3})a = 2(2-\sqrt{3})a$$
So, $AP = x = 2AD$, which means that $ADP$ is a $30-60-90$ triangle and $\angle DAP = 60^{\circ}$