Point $D$ inside $\triangle{ABC}$ such that $\angle{ABD}=\angle{CBD}=6^{\circ}$,$\angle{BCD}=12^{\circ}$,$\angle{ACD}=18^{\circ}$, $\angle{BAD}=?$

Question

Point $D$ is in $\triangle{ABC}$ such that $\angle{ABD}=\angle{CBD}=6^{\circ}$,$\angle{BCD}=12^{\circ}$,$\angle{ACD}=18^{\circ}$, $\color{red}{\angle{BAD}=?}$

Can we find a pure elementary geometric approach to answer this question?

Answer 1

The following approach should work generally to identify solutions in $6n^\circ$ adventitious triangles. Points $P,Q$ mark the triangle in question and the inner intersection in question. Within the larger context of $6n^\circ$ triangles, these are vertices of a rhombus marked with the other two pink dots (since the diagonals are perpendicular), hence the opposing angles are equal, and bisected by the diagonals, and we instantly know the relevant angle: $42^\circ$.

Proof: The image below shows the lattice again, this time marking all points that lie on $6n^\circ$ rays from all $3$ vertices. $P$ and $Q$ will lie on such point. Once this fact is established, the proof above follows.

I) By an argument first of mirror symmetry, then by $120$-degree radial symmetry, all filled gray points are on rays from each vertex measuring $6n^\circ$.

II) Points $Q$ and $R$ are also on $3$ rays, and again by mirror symmetry then radial symmetry these points populate the lattice as the hollow diamonds.

Diamond proof: That rays from each corner meet in the diamond vertices can be first guessed and then confirmed in the diagram. For $Q$ the image below shows how to construct on rays the two triangles that, when rotated (upper black clockwise $120^\circ$, pink counter-clockwise $120^\circ$), will perfectly match the triangle on the bottom to complete the equilateral (corners sum to $60^\circ$, equilateral sides, central angles sum to $360^\circ$). We can confirm that these triangles meet in a point by using the law of sines and Ceva's theorem, which yield $\frac{\sin{6}\sin{42}\sin{42}}{\sin{12}\sin{18}\sin{54}}=1$. This is the only possible pair of triangles that can meet at one point to complete the equilateral (at $Q$). The argument for $R$ is similar.

Alternate diamond proof: we can also follow synthetic proofs to the same conclusion, such as the one the OP added for $Q$ and this one for $R$: With initial triangle $ARB$ subtending $6$ and $18$ degrees, form the darts $AB_{12}RB$ and $AB_{12}SB$, note the formation of the $4$ isosceles triangles within the darts. We also have $CS$ at right angles to the base, so $\angle{CSR}=90-6=84$. But $AB=AB_{12}=AC$ so we have isosceles $\triangle{AB_{12}C}$ and $\angle{CB_{12}A}=66^\circ$, which, when combined with the angle in the large dart, also yields $84^\circ$. Thus we have the kite $CSRB_{12}$ and diagonal $CR$ bisecting $36^\circ$ at $C$, so $\angle{RCB_{12}}=18^\circ$ and within that $\angle{BCR}=12^\circ$.

The lattice above provides natural paths to solutions for virtually all $6n$ adventitious triangles. The solution added by the OP essentially follows the rhombus path, establishing the $3$-ray convergence for point $Q$ as as lemma, then using a kite instead of a rhombus, and the overall diagram makes it clear why this is a useful route toward a solution. Fun fact: Bottoma's triangle, a counterexample to "equal external angle bisectors $\Rightarrow$ isosceles" is hiding in the lattice as $\triangle{ABM}$ in the second diagram.

Answer 2

Here is the solution in pure elementary geometric approach:

\begin{multline} \shoveleft \text{First we need solve a related problem: }\\ \shoveleft \text{Let }D \text{ be a point in equilateral }\triangle{ABC} \text{ such that }\angle{BCD}=12^{\circ}, \angle{CBD}=6^{\circ}, \color{red}{\angle{CAD}=?} \end{multline}

This can be solved:

\begin{multline} \shoveleft \text{Make }E \text{ be the reflect of }C \text{ along the line }BD \implies \\ \shoveleft CD=DE,BC=BE, \angle{BCD}=\angle{BED}=6^{\circ}, \angle{BCD}=\angle{BED}=12^{\circ}\\ \shoveleft \implies \angle{BAE}=\angle{BEA}=\dfrac{180^{\circ}-\angle{ABE}}{2}=\dfrac{180^{\circ}-48^{\circ}}{2}=66^{\circ}\implies \angle{CAE}=6^{\circ}\\ \shoveleft \text{Extend }CD \text{ and intersect }BE \text{ at }F \implies \angle{BCF}=\angle{CBF}=12^{\circ}\implies \\ \shoveleft \begin{cases} BF=CF\implies \triangle{ABF}\cong\triangle{ACF}\implies \angle{CAF}=30^{\circ}\implies \angle{EAF}=36^{\circ}\\ \angle{CFE}=24^{\circ} \implies \angle{CDE}=36^{\circ}=\angle{EAF}\\ \end{cases}\\ \shoveleft \implies AFDE \text{ is cyclic }\implies \angle{DAE}=\angle{DFE}=24^{\circ}\implies \angle{DAC}=\bbox[5px, border: 1px solid black]{18^{\circ}}\\ \shoveleft \\ \end{multline}

Now we can come back to original problem:

\begin{multline} \shoveleft \text{Make }\triangle{EBC} \text{ be an equilateral triangle}\\ \shoveleft \implies \angle{DEC}=18^{\circ} \implies \angle{BED}=42^{\circ}\\ \shoveleft \text{And we know }\angle{ABE}=48^{\circ} \implies BA\perp DE, \text{ suppose }BA\cap DE=G \\ \shoveleft \text{Extend }CA \text{ to }F\text{ on }BE, \angle{BAF}=30^{\circ} \implies CF \perp BE, BF=FE\\ \shoveleft \text{Make }H \text{ on }AF \text{ such that }\angle{ABH}=6^{\circ} \implies \angle{ABH}=42^{\circ}\\ \shoveleft \triangle{BFH}\cong\triangle{EFH}\implies BH=EH\implies \angle{FEH}=42^{\circ}=\angle{BED}\\ \shoveleft \implies EHGD \text{ are collinear} \implies BD=BH \implies \triangle{ABD}\cong\triangle{ABH}\\ \shoveleft \implies \angle{BAD}=\angle{BAH}=90^{\circ}-\angle{AHG}=90^{\circ}-\angle{EHF}=\angle{FEH}=\bbox[5px, border: 1px solid black]{42^{\circ}}\\ \end{multline}

Answer 3

Here is a trigonometric solution, it leads to a different equation that you mentioned, but it is still quite difficult, using law of sines in triangles $ABC$, $ABD$ and $BCD$ we have \begin{align*} &\frac{AB}{\sin{30}} = \frac{BC}{\sin{42}} \Longrightarrow AB = \frac{BC}{2\sin{42}}\\ &\frac{BD}{\sin{x}} = \frac{AB}{\sin{(174-x)}} \Longrightarrow BD = \frac{AB\sin{x}}{\sin{(x+6)}} = \frac{BC\sin{x}}{2\sin{42}\sin{(x+6)}} \\ &\frac{BC}{\sin{18}} = \frac{BD}{\sin{12}} \Longrightarrow BC = \frac{BC\sin{x}\sin{18}}{2\sin{12}\sin{42}\sin{(x+6)}} \end{align*} which leads us to $$ \frac{\sin{(x+6)}}{\sin{x}} = \frac{\sin{18}}{2\sin{12}\sin{42}} $$ i.e. \begin{equation} \cos{6}+\cot{x}\sin{6} = \frac{\sin{18}}{2\sin{12}\sin{42}} \tag{*} \end{equation} so we have \begin{align*} \cot{x}\sin{6} &= \frac{\sin{18}}{2\sin{12}\sin{42}} - \cos{6} = \frac{\sin{18}-2\sin{12}\cos{6}\sin{42}}{2\sin{12}\sin{42}} \\ &= \frac{\sin{18}-\left(\sin{18}+\sin{6}\right)\sin{42}}{2\sin{12}\sin{42}} = \frac{\sin{18}-\sin{18}\sin{42}-\sin{6}\sin{42}}{2\sin{12}\sin{42}} \\ &=\frac{\sin{18}-\frac{1}{2}\left(\cos{24}-\cos{60}\right)-\frac{1}{2}\left(\cos{36}-\cos{48}\right)}{2\sin{12}\sin{42}} \\ &=\frac{2\sin{18}+\cos{60}-\cos{36}-\cos{24}+\cos{48}}{4\sin{12}\sin{42}} \\ &=\frac{\frac{\sqrt{5}-1}{2}+\frac{1}{2}-\frac{(\sqrt{5}+1)}{4}-\cos{24}+\cos{48}}{4\sin{12}\sin{42}} = \frac{\frac{\sqrt{5}-1}{4}-\cos{24}+\cos{48}}{4\sin{12}\sin{42}} \\ &= \frac{\cos{72}+\cos{48}-\cos{24}}{4\sin{12}\sin{42}} = \frac{2\cos{60}\cos{12}-\cos{24}}{4\sin{12}\sin{42}} = \frac{\cos{12}-\cos{24}}{4\sin{12}\sin{42}} \\ &= \frac{2\sin{18}\sin{6}}{4\sin{12}\sin{42}} = \frac{\sin{18}}{2\sin{12}\sin{42}}\sin{6}, \end{align*} now, using (*) again we have \begin{align*} &\cot{x}\sin{6} = \left(\cos{6}+\cot{x}\sin{6}\right)\sin{6} \\ &\cot{x} = \cos{6}+\cot{x}\sin{6} = \sin{84}+\cot{x}\cos{84} \\ &(1-\cos{84})\cot{x} = \sin{84} \\ &2\sin^{2}{42}\cot{x} = 2\sin{42}\cos{42} \\ &\cot{x} = \cot{42}, \end{align*} hence $x=42$.