Geometry Question on a equilateral triangle (Triangle and Square)

$\angle CAB = \angle CBA = 15$ . Draw $\Delta BED$ congruent to $\Delta ABC$ , so that $\angle EBD = \angle EBD = 15$ . Then since $\angle CBE = 60$ and $CB = EB$ , $\Delta CBE$ is equilateral triangle. Therefore $CE = BE$ and $\Delta CED$ and $\Delta BED$ are congruent. Hence $CD = DB = FD$ , so that $\Delta CDF$ is equilateral triangle with all angles $60$ .

Let $P'$ be a point such that $ABP'$ is equilateral. e.g ( $AB=AP'=BP'$ )

Then we get that $\angle P'AD = \angle P'BC = 30^{\circ}$

Since $AD=AP'=BP'=BC$ , we get that $\angle AP'D = \angle BP'C = 75^{\circ}$

We then get $\angle DP'C = 150^{\circ}$ and $\Delta P'DC \cong \Delta PDC$ and since $P$ and $P'$ lie inside the square. We get $P \equiv P'$ .

Note that we made use of the phantom point $P'$ to arrive to our conclusion.

If you assume the side of the square to be 2, then the distance between P and CD is $tan(15^\circ)=2-\sqrt3$ .

The distance of P from AB, if ABP is an equilateral triangle with side 2 is $\sqrt3$ .

The vertical distance between AB and CD, measured at P, is therefore $2-\sqrt3+\sqrt3=2$ as required.