Geometry problem about angles in triangles In the triangle $ABC$, the point $D$ in $BC$ is so that $\overline{AC}=\overline{BD}$. Let $\angle ABD = \theta$, $\angle ACD = 4\theta$ and $\angle CAD = 90^\circ-\theta$.
How can I calculate $\theta$?

I tried using Stewart's theorem on $\triangle ABC$ and the law of cosines on some angles. This resulted in a system of 5 equations, which I wasn't able to solve for $\theta$. Wolfram|Alpha says the solution is $\theta=20^\circ$. Is there a simpler way to solve this problem?
 A: 
Hints:
The given data is compatible with isosceles triangle as can be seen in figure and $\theta=20^o$. You have to show $AB=BC$.
A: Trigonometric solution: This goes straightforward in a short story. Use the sine theorem in the triangles $\Delta ABC$ and $\Delta ABD$ (for instance), then trigonometry, as mentioned in the comments. Let's go.
I will use $t$ instead of $\theta$ for easy typing.
We get
$$
\frac{\sin t}{\sin 4t} = 
\frac{AB}{AC} = 
\frac{L}{AC} = 
\frac{DC}{AC} = 
\frac{\sin(90^\circ-4t)}{\sin(90^\circ+3t)} = 
\frac{\cos 4t}{\cos 3t} \ . 
$$
The obtained equation is
$$
\begin{aligned}
% \sin t\cos 3t &=\sin 4t\cos 4t\ , &&\text{equivalently}\\ 
2\sin t\cos 3t &=2\sin 4t\cos 4t\ , &&\text{equivalently}\\ 
\sin 4t-\sin 2t &=\sin 8t\ , &&\text{equivalently}\\ 
\sin 4t-\sin 8t &=\sin 2t\ , &&\text{equivalently}\\ 
2\sin\frac{4t-8t}2\cos\frac{4t+8t}2 &= \sin 2t\ , &&\text{equivalently }(\sin 2t\ne 0)\\ 
\cos 6t &=-\frac 12\ .
\end{aligned}
$$

*

*Then $6t=120^\circ$ leads to the solution $t=20^\circ$.
We can draw the corresponding isosceles triangle $\Delta ABC$ with angles $80^\circ$, $80^\circ$, $20^\circ$. And verify geometrically that it is indeed a solution. (Sufficiency.)

*The "other", the next solution of $\cos 6t=-1/2$ is for $6t=240^\circ$, i.e. $t=40^\circ$, it has to be rejected, since $\widehat{DAC}=90^\circ-4t$ has to be $>0$ to match the figure, and insure that $D$ is on the segment $[BC]$.

So the solution is $\color{blue}{t= 20^\circ}$.
$\square$

Note: I tried hard to also give  a geometric, direct solution.
(I.e. a solution where we start with the given data, and obtain a geometric property determining the unknown $t$. All this without using "injectivity" arguments, then checking that $t=20^\circ$ is indeed a solution.) Without success.
Please stop reading  here if this kind of geometrical wish is meaninless when seen from your screen.
One try was as follows.
We work in a constellation of points as in the OP, in particular try to use the important information $AB=DC$.
Let $E$ be the reflection of $A$ w.r.t. the line $BDC$.
We construct the isosceles trapezium $EDCF$, where $F$ is taken so that $ED\|FC$ and $DC=EF$. Then there is a picture like the following one.

Related angles depending on the points $A,B,C,D,E,F$ were then computed depending on the parameter $t$.

*

*The angles in $B,C$ in $\Delta ABC$ are given, $4t$ and $t$.
We know $\widehat{BAD}=90^\circ-t$. This implies $\widehat{DAC}=90^\circ-4t$, so that the sum of the angles in $\Delta ABC$ is $180^\circ$.

*In $\Delta ADC$ we get $\widehat{ADC}=180^\circ-(90^\circ-4t)-t=90^\circ+3t$.

*So its supplement is $\widehat{ADB}=90^\circ-3t$.

*We also have the reflected angles
$\widehat{EDC}=90^\circ+3t$ and
$\widehat{EDB}=90^\circ-3t$. Since $\Delta ADE$ isosceles, its angles in $A$ and $E$ are $3t$.

*Then $\widehat{BAE}=\widehat{BEA}
=\widehat{BAD}-\widehat{EAD}=(90^\circ-t)-3t=90^\circ-4t$.

*Also, by reflexion,  $\widehat{DEC}=\widehat{DAC}=90^\circ-4t$.

*$EDCF$ is isosceles, so we know all its angles between sides and diagonals, as shown in the picture.

*We compute $\widehat{BEF}$, knowing all other angles in $E$ in the picture:
$$
\begin{aligned}
\widehat{BEF}
&=360^\circ-\widehat{BED}-\widehat{DEF}\\
&=360^\circ-\widehat{BAD}-\widehat{EDC}\\
&=360^\circ-(90^\circ-t)-(90^\circ+3t)
=180^\circ-2t\ .
\end{aligned}
$$

*(It would be nice to be able to show somehow $A,B,F$ colinear...

*We finally use the given metric relation, and obtain:
$$
BE=BA=DC=EF\ ,
$$
so $\Delta BEF$ isosceles, with angles in $B,F$ equal to $t$.

Now let $S$ be the intersection $S=EF\cap AC$. Then $\Delta BCF=\Delta SFC$, a common side $CF=FC$ and respectively equal angles adjacent to it. We get $SC=BF$. This implies $\Delta SCD=\Delta BFE$, so $BE=EF=SD=DC$. We obtain $\widehat{DSC}=t$ and $\widehat{FSD}=\widehat{CBE}=4t$. But again we are missing geometric information to get $\widehat{ASF}$. We only know $\widehat{ASF}=180^\circ-5t$, and need $\widehat{ASF}=4t$. (This is as hard as the colinearity of $A,B,F$.)
I decided to give up this path and stop searching for a solution. Well, sometimes long, unsuccessful attempts still make sense, they fail for a more or less explicit reason. In this case, we are trying to solve an issue by geometrical means which is eqivalent to solving a trigonometric equation as above. Note that for the intermediate step $\sin 8t -\sin 4t=\sin 2t$ the above picture shows the angles with measures $8t$, $4t$, $2t$, for instance in $B$ and $C$. If we norm $AB=1$, then $AE=2\sin 4t$, and we may force by construction also the other two sines. But OK, i do not have anything concrete, so i will stop here.

Note:
Strictly speaking we did not show that $t=20^\circ$ is indeed a solution to the problem. (We have only shown that with all the angle data from the figure and $AB=CD$ we then have $t=20^\circ$.) So let us check:

Let us norm $AB=1$ and show that in the picture $CD$ has the same length:
$$
\begin{aligned}
CD 
&= BC-BD
=
\frac{\sin 80^\circ}{\sin 20^\circ}
-
\frac{\sin 70^\circ}{\sin 30^\circ}
=
\frac{2\sin 40^\circ\cos 40^\circ}{\sin 20^\circ}
-
\frac{\cos 20^\circ}{1/2}
\\
&=
4\cos 20^\circ\cos 40^\circ - 2\cos 20^\circ
\\
&=2\Big[\cos (40^\circ+20^\circ) + \cos (40^\circ-20^\circ)\Big]- 2\cos 20^\circ
=2\cos 60^\circ
\\
&=1\ .
\end{aligned}
$$
$\square$
But this time the geometrical proof is simpler: Construct $E$ inside $\Delta ABC$ so that $\Delta ABE$ is equilateral. Then $C$ and $E$ are on the perpendicular bisector of $AB$, which is also angle bisector in $C$. So the angles in $\Delta AEC$ are known, the ones in $A$ and $C$ are $20^\circ$ and $10^\circ$. The "same" happens to be the case for $\Delta ADC$ (for angles in $C$ and $A$). The two triangles share the common side $AC$, so they are equal, so $AB=EA=DC$.

Bonus: From $\Delta ADC=\Delta AEC$ we easily get $AEDC$ an isosceles trapezium, then the angles in $C,D$ in $\Delta CDE$ ($10^\circ$ and $160^\circ$), so the missing angle is $10^\circ$, so this triangle is isosceles, finally giving:
$$
AB=BE=AE=DE=CD\ .
$$
