(Solved)
I have this problem:
In the figure we have a square $ABCD$ with diagonal $BD$ and the segment $AM$ such that $M$ belongs to the side $BC$ and such that $\overline{MC} = \overline{MP}$ where $P$ is the intersection of $AM$ with $BD$. Find the measure of $\angle BMP = \alpha$.
So I tried a little bit. I drew on Geogebra and I´ve $\alpha = 60^{\circ}$. But I could not do this. I know that $\Delta ADP$ is the similarity of $\Delta MBP$ because of $\angle DAM = \alpha$. I considered $d(A,D) = a$ and $d(M,C) = y$ and this implies $d(B,M) = a - y$. With this, I tried a lot of relations using law of Sinus and law of Cosines and some corresponded with this similarity. I don´t have success. I tried to use Analytic Geometric too. I considered $A=(0,0)$, $D=(a,0)$, $C=(a,a)$, $B=(0,a)$ and $M=(a-y,a)$ I used the equation $y = -x +a$ that contains $BD$ and I know that I have a circumference of centre M and radius $MC$. But I can´t find the angle $\alpha$.
(Solved). Thanks too much.
Let $AB=a$ and $PM=MC=b$.
Thus, since $\Delta BMP\sim\Delta DAP,$ we obtain: $$\frac{BM}{AD}=\frac{MP}{AP}$$ or $$\frac{a-b}{a}=\frac{b}{\sqrt{a^2+(a-b)^2}-b}$$ or $$(a-b)\sqrt{2a^2-2ab+b^2}=2ab-b^2$$ or $$(b^2-2ab+a^2)(b^2-2ab+2a^2)=(b^2-2ab)^2$$ or $$(b^2-2ab)^2+3a^2(b^2-2ab)+2a^4=(b^2-2ab)^2$$ or $$3b^2-6ab+2a^2=0$$ or $$b=\frac{3a-\sqrt{3a^2}}{3}$$ or $$b=a-\frac{a}{\sqrt3}.$$ Thus, $$BM=\frac{a}{\sqrt3}$$ and $$\tan\alpha=\frac{a}{\frac{a}{\sqrt3}}=\sqrt3.$$ Can you end it now?