Is there a simpler geometric solution?

Question

In the diagram below, $ABC$ is a triangle. Given that $\overline{AD}=\overline{BC}$, $\angle ABC=120^{\circ}$, $\angle BDA=3\phi$, and $\angle BCA=2\phi$, determine the measure of $\phi$.

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Construct the equilateral triangle $BQC$ and the parallelogram $ABPD$. Angle chasing gives $\angle DBC = \phi$, $\angle DAB\cong PBQ \cong BPD=60^{\circ}-2\phi$, $\angle BDP = 120^{\circ}-\phi$. $\triangle BPQ$ is isosceles, thus $\angle BQP\cong\angle BPQ=60^{\circ}+\phi$. It follows that $BQPD$ is cyclic. It follows that $\angle BPD=60^{\circ}-2\phi$, thus $\angle BQD=60^{\circ}-2\phi$. It follows that $\triangle BQD$ is isosceles, and so is $\triangle DQC$. It follows that $\angle DBQ\cong\angle QDB = 60^{\circ}+\phi$. It follows that the angles of the triangle $DQC$ is $60^{\circ}+2\phi+60^{\circ}+2\phi+2\phi$. Thus $\phi=10^{\circ}$.

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It feels as if there should be a simpler geometric solution. Can you come up with one?

Answer 1

[I like your synthetic solution. One could always brute force it out with trigo.]

From sine rule on ABD, $ \frac{ BA} { \sin 3 \phi} = \frac{AD}{ \sin (120^\circ - \phi)}$

From sine rule on BAC, $\frac{BA}{ \sin 2 \phi} = \frac{BC} { \sin (60 ^ \circ - 2 \phi)}$

Hence, $\frac{ \sin 3 \phi} { \sin (120^\circ - \phi) } = \frac{BA}{AD} = \frac{BA}{BC} = \frac { \sin 2 \phi} { \sin ( 60 ^ \circ - 2 \phi )} $

Expanding this and solving for $\phi$, bearing in mind that $\phi < 30 ^ \circ $ from geometric considerations, gives us $\phi = 10 ^ \circ$.

Answer 2

Construct a point $D'$ inside $\triangle ADB$, such that $\triangle DCB\cong \triangle D'AD$, which is possible since $\vert AD\vert=\vert CB\vert$. Now reflect $D'$ along $AD$ in order to obtain $D''$ which, again, leads to $\triangle ADD''\cong \triangle ADD'\cong\triangle DCB$.

Observe now that $$\triangle D''DD'\cong \triangle D'DB\implies \color{green}{2\cdot \vert D'E\vert=\vert D'B\vert}\tag{1}$$ Angle chasing in $\triangle ACB$ shows $$\angle CAB=60°-2\phi\implies \angle D'AB=60°-2\phi-\angle DAD'=60°-4\phi\tag{2}$$ Thus $\angle D''AB=60°$. Furthermore, angle chasing confirms that $$\angle DBC=\angle BDA-\angle BCD=\phi\qquad \angle D'BD=\frac{180°-\angle BDD'}{2}=\frac{180°-2\phi}2=90-\phi$$ Putting this together yields $$\angle ABD'=120°-\angle D'BD-\angle DBC=30°$$ If we extend $BD'$ to intersect $AD''$ at $F$, we notice that $\triangle AFB$ is a $30°-60°-90°$ triangle.

Let $G$ be the point on $AB$ such that $\angle D'GB=90°$. Now, $\triangle BGD'\sim \triangle AFB$. Hence $$\color{green}{2\cdot \vert GD'\vert=\vert D'B\vert\stackrel{(1)}{=}2\cdot\vert D'E\vert\implies \vert GD'\vert=\vert D'E\vert}$$ And we are almost done, since this implies that $AD'$ bisects $\angle DAB$. Therefore, and finally $$2\phi=\angle DAD'=D'AB\stackrel{(2)}{=}60°-4\phi\iff 2\phi=60°-4\phi\iff\fbox{$\ \phi=10°\ $}$$

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Sidenote $\ $ I don't know if this solution is simpler than the one you've offered. What I do know, is that this proof is almost elementary after one has introduced all points I use and has performed angle chasing...