Let $ABCD $ a square and $P $ inside the square s.t. $\angle PCD=\angle PDC=15^{°} $. Let $ABCD $ a square  and $P $ inside the square s.t. $\angle PCD=\angle PDC=15^{°} $. Show that $PAB $ is equilateral. 
It's easy to prove that $PA=PB $. But I need a construction to show that the others sides are congruent.
It's possible to solve it without trigonometry? I need a construction.
 A: 
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A: Long story short (i.e. assume the side-length is 2 and bisect the square through $P$), you want to show that $\tan15^\circ+\tan60^\circ=2$.  That's not hard to do with the half-angle formula.
As you need to verify this solution with a construction, let's do that:

Your easiest bet is to construct a square, then identify $P$ by swinging arcs centered at $A$ and $B$ with radius $AB$.  Clearly, $m\angle PAB=60^\circ$.  Now, bisect $\angle PAB$, then bisect that angle to create an angle with measure $15^\circ$.  Copy that angle to verify that it coincides with $\angle CDP$ and you've shown that $m\angle CDP=15^\circ$.
A: 
Extend DP to meet CB at E. The area of the triangle DCE is
$$\frac 12 DC\cdot CE = \frac 12 DE\cdot \frac {PC}{2}$$
With PC = $\frac12$DE and CD = BC, we have
$$\frac {CE}{PC} = \frac{PC}{CB}$$
Therefore, the triangles PCE and BCP are similar. Since the triangle CPE is isosceles, so is the triangle BCP. Thus, CB = BP and the triangle ABP is equilateral. 
