What's the measure of angle $x$ in the triangle below? 
Any point $D$, which is interior to the triangle $\triangle ABC$, determines vertex angles
having the following measures:
$m(BAD) = x,$
$m(ABD) = 2x,$
$m(BCD) = 3x,$
$m(ACD)= 4x,$
$m(DBC) = 5x.\\$
Find the measure of $x$.
(Answer:$10^\circ$)


My progress:
$\dfrac{AB}{AD}=\dfrac{\sin(180−3x)}{\sin(2x)}\\
\dfrac{AC}{AD}=\dfrac{\sin(11x)}{\sin(4x)}
AB=AC \implies 
\dfrac{\sin(3x)}{\sin(2x)}=\dfrac{\sin(11x)}{\sin(4x)}$
but it is a complex equation to solve. Is there another way?



Let's do this the other way around.
Assume a triangle $\triangle A'B'C'$ with $\angle B'A'C'=40^{\circ}$ and $A'B'=A'C'$. Let $D'$ be an interior point of $\triangle A'B'C'$ such that $\angle B'A'D'=10^{\circ}$ and $\triangle A'B'D'=20^{\circ}$.
We construct $\triangle A'D'O\cong \triangle A'D'B'$ as shown in the figure above. Then $$\angle B'D'O=
\angle B'A'D' + \angle OA'D' + \angle A'B'D' + \angle A'OD'=60^{\circ},$$ implying that $\triangle B'D'O$ is equilateral, so $B'D'=OD'=B'O$. We also obtain $\angle C'B'O=10^{\circ}$.
Since $\angle B'A'O=20^{\circ}=\frac{1}2\angle B'A'C'$, $A'O$ bisects $\angle B'A'C'$. This means $\triangle B'A'O\cong \triangle C'A'O$, so $B'O=C'O$. By symmetry, $\angle B'C'O=\angle C'B'O=10^{\circ}$.
Now we have $B'O=D'O=C'O$, and thus $O$ is the circumcenter of $\triangle B'C'D'$. This implies that $$\angle C'OD'=2\angle C'B'D'=100^{\circ}$$ yielding $$\angle OD'C'=\angle OC'D'=40^{\circ}.$$ Hence $$\angle B'C'D'=\angle OC'D'-\angle B'C'O=30^{\circ}$$ and $$\angle A'C'D'=70^{\circ}-\angle B'C'D'=40^{\circ}.$$
We conclude that $A'\equiv A$, $B'\equiv B$, $C'\equiv C$ and $D'\equiv D$, and $\color{red}{x=10^{\circ}}$ will be our final answer.
