Square-To what should $x$ be equal? 
At the side $AD=a$ of the square $ABCD$, we take the point $E$ and at the extension of $AB$ the point $Z$, such that $DE=BZ=x$.
a) Show that the line bisector of $EZ$ goes through the vertex $C$.
b) To what should $x$ be equal, so that the line bisector goes through the midpoint of $AB$?
c) To what should $x$ be equal, so that $(CAM)=(EAZ)$?
$$$$
My attempt:
c) $(EAZ)=\frac{1}{2}(a+x)(a-x)=\frac{1}{2}(a^2-x^2)$
From the Pythagorean Theorem: $(AD)^2+(DC)^2=(AC)^2 \Rightarrow (AC)^2=2a^2 \Rightarrow (AC)=a \sqrt{2}$
$(CAM)=\frac{1}{2}a \sqrt{2} \cdot (\text{ height })$
Which is the height of this triangle?
a) Do I have to show that $(EC)=(CZ)$? Or do I have to show that $(CM)$ is perpendicular to $(EZ)$?
I have no attempt for b).
Any help would be appreciated!
 A: a) Yes, showing that EC=CZ would do the job.
b) If the bisector goes through the midpoint of AB (call it K) then the gradient of KC is $a/(a/2)=2$. The gradient of EZ is $(x-a)/(a+x)$. Perpendicular gradients $m_1$ and $m_2$ satisfy $m_1m_2=-1$ so this allows you to find $x$.
c) Note that the length of EAZ is $2a$ irrespective of $x$. So you just need to find the $x$ that gives CAM=$2a$.
A: Let $AD = DC=CB=BA=a$ and $DE =BZ=x$.
a) 
At first note that the triangles $\triangle CDE$ and $ \triangle CBZ$ are congruent.
Hence we can conclude that $CE = CZ$.
As $EM =MZ$ and $CE=CZ$ it follows that $CM$ is the perpendicular bisector of segment $EZ$.
b)
Let $K$ the intersection point between the lines $CM$ and $AB$. 
The triangles $\triangle EAZ$ and $\triangle KBC$ are similar.
Recalling that $K$ is also the midpoint of segment $AB$ we get:
$$\frac{a-x}{a+x}=\frac{\frac{a}{2}}{a}.$$
Therefore
$$x=\frac{a}{3}.$$
c)
Note that$(AMC)=(AMCD)-(ADC)$.
But $(AMCD)=2(AMD)$(Can you prove that $AM =MC$?).
Hence
$$(CAM)=2\frac{a(a+x)}{2}-\frac{a^2}{2}$$
$$ \Rightarrow (CAM)=\frac{a x}{2} \quad (1) $$
The area of $\triangle EAZ$ can be calculated by:
$$(EAZ)= \frac{(a-x)(a+x)}{2} \quad(2) $$
As equations $(1)$ and $(2)$ must be equal, we get:
$$x^2+ax-a^2=0 \quad(3)$$
Solving $(3)$ we get:
$$x= \frac{a(\sqrt{5}-1)}{2}$$
