How to find the measure of an angle on the interior of an isosceles triangle? The problem is as follows:

On the interior of an isosceles triangle $\triangle{ABC}$ where $\angle{B}=110^{\circ}$ it is situated a point $M$ such as $AB=MC$ and $\angle{BAM}=5^{\circ}$. Using this information find $\angle{MCA}$.

The choices in my book are as follows:
$\begin{array}{cc}
1.&10^{\circ}\\
2.&15^{\circ}\\
3.&20^{\circ}\\
4.&25^{\circ}\\
5.&30^{\circ}\\
\end{array}$
According to the official answers sheet the answer is choice 4. But how would you get there?.
I've been looking at this figure and I am out of ideas. Can someone help me with a sketch for this problem and how to solve it?.
This problem should be solved relying only in euclidean geometry constructions, is there a way to do that to solve this?. Since I am not good with that I will really appreciate someone could help me here.
 A: 
Idea: I feel that the problem gives $m(\widehat{MAB})=5^{\circ}$, not $m(\widehat{MAC})=30^{\circ}$ to get us out of the way. Then, to use this angle, I have to get equilateral or right triangles somewhere. In order to use it effectively, I can prefer a circle around it.
Solution: Let $O$ be the center of the circle of circles of the triangle $MAC .$ $\\$ Since $|CO|=|MO|=|AO|$ and $2m(\widehat{MAC})=m(\widehat{MOC})=60^{\circ}$(relationship between central angle and inscribed angle), $CMO$ is an equilateral triangle. So $m(\widehat{OCA})=m(\widehat{CAO})=60^{\circ}-m(\widehat{MCA}).$ Now, let's draw perpendicular to $[CA]$ from $O$ and call the point $F$ where the drawn line intersects $[CA].$ Since $|CB|=|BA|$ and $CO|=|AO|$, $|CF|=|FA|$ passes through $[OF$] or rather $B$ (Originally this $ABCO$ is a property of a rhombus, the diagonals intersect perpendicularly). Where $m(\widehat{CBO})=m(\widehat{OBA})=55^{\circ}$ and $|BA|=|AO|$ we get $m(\widehat{BAO})=70^{\circ}$ using $m(\widehat{OBA})=55^{\circ}$ and $|BA|=|AO|$ we get $m(\widehat{BAO})=70^{\circ}$. Now, the problem is solved.
$$m(\widehat{CAO})=60^{\circ}-m(\widehat{MCA})=70^{\circ}-35^{\circ}$$
$$m(\widehat{CAO})=60^{\circ}-m(\widehat{MCA})=35^{\circ}$$
$$m(\widehat{MCA})=25^{\circ} \therefore$$
A: Let $M'$ be a point inside $\triangle ABC$, such that $BC = M'C$, and $\angle BCM' = 10^\circ$. We shall show that $M' = M$. Let $A'$ be the reflection of $A$ over $BM'$.
Since $\triangle BCM'$ is isosceles, we get $\angle M'BC = 85^\circ$. So, $\angle M'BA = 25^\circ = \angle M'BA'$. Therefore, $\angle A'BC = 85^\circ - 25^\circ = 60^\circ$. Moreover, $BC = BA = BA'$, so $\triangle A'BC$ is equilateral. So, $CA' = CB = CM'$. Therefore, $C$ is the circumcenter of $\triangle A'M'B$. Now, degree measure theorem yields $\angle BAM' = \angle BA'M' = \frac12 \angle BCM' = 5^\circ$.
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So, $\angle BAM' = 5^\circ$, and $BC = M'C = AB$, and this characterizes the point $M'$. So, $M' = M$, and thus, $\angle MCA = \angle M'CA = 35^\circ - 10^\circ = 25^\circ$.
A: Here is a further solution based on the symmetries of a regular polygon. In this case, a regular $36$-gon $0123\dots$, needed to cover the $5^\circ$ angle from the problem. The prime-notation means reflection w.r.t. the symmetry line $0\ 18$. We start with the following picture, which realizes the given $\Delta ABC$ as $\Delta 707'$ with vertices among those of a $36$-gon:

Claim: The chords $1'7$, $0\;12$, $1\; 14'$, $27'$ are concurrent in a point $M$.
Proof: The angle bisectors in $\Delta 0\, 2\, 14'$ are concurrent, these are the last three chords in the list. Let us denote by $M$ this incenter.
It is also the orthocenter of $\Delta 1\, 7'\,12$, built from the other three vertices involved in the chords.

(Why does $1'7$ go through this point, too?! The simplest solution i found also constructs the point from the other solutions, but in a different setting with a different argument.)
Angles between chords in the picture are easily computed,
for instance the angle
$\widehat{0M7'}$
between $27'$ and $0\;12$ is the mean of the arcs measured they delimit, $\frac 12((0'-7')+(12-2))\cdot10^\circ=85^\circ$.
This is the same measure as for $\widehat{M07'}$, so $\Delta 7'0M$ isosceles.
Let now $M^*$ be the reflection of $M$ w.r.t. $77'$. Since $M\in 7'2$, the reflection $M^*$ is on the reflected line $7'12$. We have $7'M^*=7'M=7'0$, and the angle $\widehat{07'M^*}=\widehat{07'12}=\frac 12\cdot (12-0)\cdot 10^\circ=60^\circ$. So $\Delta 07'M^*$ is equilateral:
$$
\tag{$\dagger$}
0M^*=7'M^*=7'M=7'0=70\ ,
$$
and $M^*$ is also on $0\; 17$.
We compare now the triangles
$$
\begin{aligned}
&\Delta 0MM^* &&\text{ and}\\
&\Delta 0M7 \ .\\
\end{aligned}
$$
$07$ is a common side, $0M=0\;12$ is the angle bisector of
$\widehat{70M^*}=\widehat{7\;0\;17}$, given two congruent angles in $0$, and $(\dagger)$ gives a third side, $0M^*=07$. We compute now
$$
\widehat{M70} =
\widehat{MM^*0} =
\widehat{M^*\;0\; 18} =
\widehat{17\;0\; 18} =
5^\circ\ ,
$$
so $M$ is also on the chord $71'$, the one building a $5^\circ$ degree angle with $70$ inside $\Delta 707'=\Delta ABC$.
This finishes the proof.
$\square$

Note: From $\Delta 0MM^* \cong\Delta 0M7$ we obtain $M7=MM*$, so $\Delta 7MM^*$ is isosceles in $M$, which together with the knowledge of
$\widehat{7MM^*}=2\cdot\widehat{7\;M\;12}=2\cdot 30^\circ=60^\circ$ makes it equilateral. We get $7,M^*,13'$ colinear. As already observed, $M^*\in 7'12$, and it seems that $M^*$ is also on a fourth chord, $14'6$.
