anglegeometrytriangles
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?
Continue by solving the equation
$$\frac {\sin(\theta+20^\circ)} {\sin\theta}=\frac {1} {2\sin 10^\circ}.$$
as follows
\begin{align} \cot \theta & = \frac1{2\sin 10^\circ\sin 20^\circ}-\cot 20^\circ = \frac{\sin(10^\circ+ 20^\circ)}{\sin 10^\circ\sin 20^\circ}-\cot 20^\circ =\cot 10^\circ \end{align}
Thus, $\theta = 10^\circ$.
We have $\angle BAD=3x=\angle ADB$, hence $AB=BD$.
Let $E$ be the point on $AC$ such that $BE=AB$. Then $A,D,E$ lie on a circle with center $B$, hence $\angle EBD =2\angle EAD = 2x$.
On the other hand, $\angle AEB=\angle BAE=2x$, hence $\angle CBE=\angle AEB-\angle ACB=2x-x=x$.
We see that $\triangle CEB \sim \triangle CDA$ because these triangles have equal angles. In particular $\frac{CE}{CD}=\frac{CB}{CA}$. Since $\angle ACB=x=\angle DCE$, we have $\triangle ACB \sim\triangle DCE$ by SAS. Hence $\angle EDC =\angle BAC=2x$.
So $\angle DEA=\angle EDC+\angle DCE=2x+x=3x$. So $\angle DEB=\angle DEA+\angle AEB =3x+2x=5x$. Also $\angle BDE =\angle DEB=5x$ because $BD=BE$. We see now that the sum of angles of triangle $EBD$ equals $12x$. This leads to $12x=180^\circ$, hence $x=15^\circ$.
Best Answer
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 $$ \begin{aligned} % \sin t\cos 3t &=\sin 4t\cos 4t\ , &&\text{equivalently}\\ 2\sin t\cos 3t &=2\sin 4t\cos 4t\ , &&\text{equivalently}\\ \sin 4t-\sin 2t &=\sin 8t\ , &&\text{equivalently}\\ \sin 4t-\sin 8t &=\sin 2t\ , &&\text{equivalently}\\ 2\sin\frac{4t-8t}2\cos\frac{4t+8t}2 &= \sin 2t\ , &&\text{equivalently }(\sin 2t\ne 0)\\ \cos 6t &=-\frac 12\ . \end{aligned} $$
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 $t$.
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: $$ AB=BE=AE=DE=CD\ . $$