How to find angle inside a right isosceles triangle with one side and an angle given The problem is as follows:
The picture from below shows a triangle. The conditions are stated in the diagram.

The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&30^{\circ}\\
2.&15^{\circ}\\
3.&25^{\circ}\\
4.&18^{\circ}\\
\end{array}$
I'm not sure exactly what to do with this problem. So far the only thing which I could spot on was that the right triangle is $45-45-90$ but other than that I'm stuck. Can somebody guide me in the right direction with this problem?.
 A: 
Let ∠DAC = ∠DCB = $x$ and ∠DCA = ∠DAB = $y$. Then, $x+y = 45$ due to the isosceles right triangle ABC and ∠ADC = 180 - ($x+y$) = 135. Apply the sine rule to the triangle ADC,
$$\frac{\sin x}{\sin ∠ADC} = \frac{DC}{AC} = \frac1{\sqrt2}$$
which leads to $\sin x = \frac1{\sqrt2}\cdot \sin 135 = \frac 12$. Thus, ∠DCB = $x$ = 30.
A: Here is a solution using only synthetic geometry. The following figure is the "solution by picture":

$D$ is on the two emphasized circles, $\Delta XCD$ is equilateral, so  $\widehat{DCB}=30^\circ$.
$\square$

Details: (If really needed.) Let $X$, be the fourth point of the square with vertices $A,B,C$. 


*

*The given conditions conditions describe an isosceles triangle $\Delta ABC$ with a right angle in $B$, 

*and the point $\color{brown}D$ is at the intersection


*

*of the circle $(C)$ centered in $C$ through $B$ (and $X$), since $CB=CD$,

*with the circle $(X)$ centered in $X$ through $A,C$. This is because $\widehat{DAC}=\widehat{BCD}$ implies $$
\begin{aligned}
\widehat{ADC}&=180^\circ-\widehat{DAC}-\widehat{DAC}
\\&=180^\circ-\widehat{BCD}-\widehat{DAC}   
\\&=180^\circ-\widehat{BCA}   
\\&=180^\circ-45^\circ=135^\circ\ .
\end{aligned}$$ (A point on the (small) arc $\overset \frown{AC}$ of the circle $(X)$ has the property that $\widehat{ADC}$ has half of the measure of the (big) arc arc $\overset \frown{CA}$, with measure $360^\circ-\widehat{AXC} =
360^\circ-90^\circ=270^\circ$. So it measures $135^\circ$.)



Because of $CD=DX=XC$, the triangle $\Delta CDX$ is equilateral, we obtain an angle of $60^\circ$ in $C$, so 
$$
\widehat{DCB}=
\widehat{XCB}-
\widehat{XCD}=
 90^\circ-60^\circ=30^\circ\ .
$$
$\square$

Bonus: 
Let us also draw the other quarter circles centered in the other vertices of the square $ABCX$, each having radius $AB=BC=CX=XA$. By the symmetry of the picture, we obtain as in the picture intersections $D,E,F,G$, which are forming a square. 
With the arguments above, 


*

*$ABCX$, $DEFG$ are squares, 

*and the triangles
$\Delta CDX$,
$\Delta XEA$,
$\Delta AFB$,
$\Delta BGC$;
$\Delta AGD$,
$\Delta BDE$,
$\Delta CEF$,
$\Delta XFG$ are equilateral.



Later EDIT:
A simplified picture was added:

(This picture is suited for a straightforward proof for the posted problem, but it hides the symmetry in the picture. Note that many such problems arise as particular, partial pictures in a constellation involving a regular polygon. Finding it is a good way to understand the special situation, and to be able to react in similar situations. Or to even compose problems.)
