# Is there a simpler geometric solution?

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 $$\phi$$.

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 $$\phi=10^{\circ}$$.

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

• @Piquito Are you saying either of those conditions is enough to solve the problem? How? – blackened May 12 at 6:10
• I think I made a mistake. I'm sorry – Piquito May 13 at 16:46
• Assuming $\overline{AD}=\overline{BC}$ one has the relation between $\theta=\angle ABC,\alpha=\angle BCA$ and $\beta=\angle BDA$ $$\frac{\sin(\alpha+\theta)}{\sin(\theta)}=\frac{\sin(\beta)}{\sin(\beta)+\sin(\beta-\alpha)}$$ and if we take $2\alpha$ and $3\alpha$ instead of $\alpha$ and $\beta$ respectively we get $$\frac{\sin(2\alpha+\theta)}{\sin(\theta)}=\frac{\sin(3\alpha)}{\sin(3\alpha)+\sin(\alpha)}$$ Consequently a value of $\theta$ in the ABC triangle corresponds to a given value of $\alpha$. Your problem was built knowing in advance the two values of $10$ and $120$ degrees. – Piquito May 13 at 17:14

[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 $$\phi = 10 ^ \circ$$.

• A GEOMETRIC solution is required by the O.P. – Piquito May 10 at 15:59
• Trigonometry is geometry, but I'd let OP decide. If he intends for a "synthetic geometry" approach, I would be somewhat surprised if there's one that's significantly better than his. – Calvin Lin May 10 at 16:01
• @Calvin Why would you be surprised? I've seen countless tough-looking geometry problems which were solved with a brilliant idea almost in a single (explanatory) line (along with an accompanying diagram). – blackened May 10 at 16:09
• Convention is that geometric $\ne$ trigonometric. Regards. – Piquito May 10 at 16:09
• @Blackened Mainly because the construction of Q is extremely motivated by the setup, and that somewhat requires P to push through any of the following details. I tried working with Q without P, but wasn't able to show (say) that QB, QD, QC had the same length. – Calvin Lin May 10 at 16:13

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 $$2\phi=\angle DAD'=D'AB\stackrel{(2)}{=}60°-4\phi\iff 2\phi=60°-4\phi\iff\fbox{\ \phi=10°\ }$$

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...

• If you are forbidden to use trigonometry, this problem is not trivial, believe me (at least for the fact of having put you an upvote) You can verify what I say giving this problem to your friends knowing maths. – Piquito May 11 at 20:05
• @Dr. The problem with “... this proof is almost trivial after one has introduced all points I use ...” is that in many (if not most) geometry solutions, once the auxiliary construction(s) are made, the rest follows trivially. I would think, coming up with that point D trick is anything but trivial. – blackened May 12 at 14:35