contest-matheuclidean-geometrygeometrytrianglestrigonometry
As title implies, the goal is to find the measure of the missing angle in the following figure. While I have solved this, which I'll show below as an answer, I'm not quite sure if my answer is accurate, so I'm posting this here to see if my answer is correct and if there are any alternative approaches. Please post your own approaches to this!
I think my solution is quite simple. Since $\triangle BDC$ is isosceles, it follows that $\angle BCD = \angle CBD = 2 \alpha$, and as such, $\angle BCA = 3\alpha$. We then compute that $$ \angle ABD = 180 - \angle CBD - \angle BCA - \angle BAC = 180 - 2\alpha - 3\alpha - 7\alpha = 180 - 12 \alpha. $$ Because also $\triangle ABD$ is isosceles, we decude from this that $$ \angle BAD = \frac{180 - \angle ABD}{2} = 6\alpha. $$ In particular, we find that $\angle CAD = 7\alpha - 6\alpha = \alpha$. This implies that also $\triangle CDA$ is isosceles; whence $|AD| = |DC| = |DB| = |BA|$. It follows that $\triangle BDA$ is even equilateral and as such, its angles are all $60^{\circ}$. Hence $6\alpha = 60^{\circ}$, and thus $\alpha = 10^{\circ}$.
This is my own approach. I'll add a brief explanation too:
This is the procedure I followed:
1.) Label the triangle $ABC$ and mark all the appropriate angles and side lengths. Notice that since $\triangle ABC$ is obtuse, its circumcenter must lie outside of the triangle itself. Locate circumcenter of $\triangle ABC$ outside and call it point $E$. Connect all the vertices of $\triangle ABC$ to $E$ via $AE$, $BE$ and $CE$. Notice that this means $AB=BE=AE=CE=CD$ because $\triangle ABE$ is equilateral.
(for further explanation of above, point $E$ is a circumcenter of $\triangle ABC$, in that case, $\angle AEB$ must be twice the measure of $\angle ACB$ (inscribed angle theorem), which means it'll be $60$. Because $\angle AEB=60$ and know that $AE=BE$, it follows that $\triangle ABE$ is equilateral)
2.) Notice also that $\angle EAC=\angle ECA=6$. Connect point $D$ with $E$ via segment $DE$. Notice that because segment $CD=CE$ and $\angle ECD=36$, this implies that $\angle EDC=\angle DEC=72$. However, notice that $\angle EBD=96-60=36$, therefore, $\angle BED$ is also $36$. But this implies that segment $BD=DE$
3.) Lastly, notice that this means $\triangle ABD$ is congruent to $\triangle AED$ via the SAS property. This means $\angle BAD=\angle EAD=x$. This means that $2x=60$, therefore, $x=30$.
Best Answer
In the figure, $\Delta AED$ together point $C$ constitute the given figure.
The constructions are:
(1) Equilateral triangle $BAC$,
(2) A circle with center $B$ and radius $BA$,
(3) Note that the circle passes through $E$ because $\angle AEC=\frac{\angle ABC}{2}$
(4) Join $DB$
Note that we do not assume that $D, E, B$ are collinear.
We are going to $\color{red}{prove}$ that $\color{red}{D, E, B}$ are $\color{red}{collinear}$.
$\color{red}{Proof:}$
Note that $\angle EAB=45^o$ and $BA=BE \implies \angle ABE=90^o$
i.e. $\color{red}{BE \bot BA}$
On the other hand, $AC=CB=CD \implies \angle ABD=90^o$
i.e. $\color{red}{BD \bot BA}$
The $2$ statements implies that $B, E, D$ are collinear.
Since $CB=CD, \angle BCA=60^o $ and $BCD$ is a trianlge, $\angle CDB=30^0$