Let $ABCD $ a square and $P $ inside the square s.t. $\angle PCD=\angle PDC=15^{°} $.

Question

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.

Answer 1

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Answer 2

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 $m\angle CDP=15^\circ$.

Answer 3

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.