Geometrical construction problem (2) There are given: $[BC],[DE]$, the angles $\phi_1$, $\phi_2$ and a point $A$. Build a  isosceles triangle $MAN$ (with$\angle{MAN}$ = $\pi/2$) with the apex of the right
angle in $A$ such as $\angle{BNC}$ =$\angle\phi_1$ and $\angle{EMD}$
=$\angle\phi_2$
thanks for your time and help! 
 A: Problem
Given the points $A$, $B$, $C$, $D$ and $E$, the line segments $BC$ and $DE$, and the angles whose measures are $\phi_1$ and $\phi_2$ ( see fig.1), find out the points $N$ and $M$ such that $m(\angle BNC) = \phi_1$, $m(\angle EMD)=\phi_2$, $m(\angle NAM) = \frac{\pi}{2}$ and $AN = AM$.

Solution
At first let’s recall the definition of a pair of arcos capazes:
A Pair of arcos capazes (Portuguese) is the geometrical locus of the points in the plane from which a given segment is seen under a certain angle.
For example in fig.2 is drawn a pair of arcos capazes of the line segment $DE$ under the angle whose measure is $\phi_2$.

The construction of a pair of arcos capazes is not difficult. See an example here:
http://fr.wikipedia.org/wiki/Arc_capable
So the solution of the original problem is (See fig.4):


*

*Draw a pair of arcos capazes $\Gamma_1$ of line segment BC under an angle whose measure is $\phi_1$.

*Draw a pair of arcos capazes $\Gamma_2$ of line segment $DE$ under an angle whose measure is $\phi_2$.

*Rotate the locus $\Gamma_2$ $\frac{\pi}{2}$ rad counter-clockwise about point $A$. You will get a new locus $\Gamma_2'$.

*Find out the point $N$, such that $N={\Gamma_1 \cap \Gamma_2’}$.

*Find out the point $M$, $M \in \Gamma_2$(rotate point N  $\frac{\pi}{2}$ rad clockwise about $A$).

*Draw the right angled isosceles $\triangle MAN$. (In our case we have two solutions $\triangle M_1AN_1$ and $\triangle M_2AN_2$).

Explanation of Steps 3 and 4. (See fig.3).
Let the isosceles triangle $MAN$ right-angled at $A$ , such that $M$ moves on $\Gamma_2$ and $A$ is fixed.
As the point $M$ moves on $\Gamma_2$ point N describes $\Gamma_2'$ which is nothing more than a copy of $\Gamma_2$ rotated $90$ degrees counter-clockwise relative to point $A$.
So the point $N$ must be the intersection point between $\Gamma_1$ and $\Gamma_2'$.
