# Exploring a Sangaku problem: proving a dilated circle is circumcircle

$$\Delta ABC \text{ is an equilateral triangle with } D \text{ being the midpoint of } BC \text{. } \Delta DEF \text{ is also an } \\ \text{equilateral triangle such that } E, F \text{ are on minor arc } BC \text{ of the circumcircle of } \Delta ABC \text{ with } \\ DE \parallel AB \text{ & } DF \parallel AC \text{. } \Delta BDG \text{ is equilateral } \text{such that } E \text{ lies on } DG\text{. } \text{ Let } H \text{ be the point on the } \\ \text{circumcircle of } \Delta DEF \text{ such that } HF \text{ is the diameter} \text{. Prove that } H \text{ is the circumcenter of } \Delta BGF.$$

The original Sangaku problem is to find the ratio of the sides of triangles ABC and DEF. It is not hard. I already have one using Pythagoras etc essentially summarized here (pastebin image). The ratio is curiously twice the golden ratio. I got stuck with this configuration looking for some novel solutions making that coincidence more obvious

EDIT So here's more context. I wish to arrive at the golden ratio bit via this, not prove that first and then this result as a corollary. How? So if we prove this, $$\text{ Now we know that } G \text{ lies on a 2x dilation } \text{ of the circumcircle of } \Delta DEF \text{ with dilation center } F \text{. }\\ \therefore FG \text{ meets the circumcircle of } \Delta DEF \text{ at } G' \text{ which bisects } FG \text{. Denote midpoint of } EF \text{ by } E'\text{. }\\ \text{We have by midpoint theorem, } E'G' \parallel EG \text{ and } EG = 2E'G'. \text{ If we call the process of going }\\ \text{ from } \Delta ABC \text{ to } \Delta DEF \textit{ shrinking,} \text{ then } G' \text{ is the vertex of the triangle on }\textit{shrinking }\Delta DEF.\\ \text{If we denote length of the edge of this twice }\textit{shrunken}\text{ triangle i.e. } \overline{E'G'} \text{ by } x \text{ and reciprocal}\\ \text{factor of }\textit{shrinking}\text{ by } 2\cdot \phi > 1\text{, then we have:}\\ \overline{EG}=2\cdot\overline{E'G'} = 2x\text{, }\quad \overline{DE} = x\cdot 2\phi\text{, }\quad \overline{DG} = \frac{\overline{BC}}{2} = \frac{x\cdot 4\phi^2}{2} = x\cdot 2\phi^2\\ \overline{DG} = \overline{DE} + \overline{EG} \implies \phi^2 = \phi + 1$$

Thus the ratio was twice the golden ratio

• FYI: The original Sangaku amounts to Odom's constructon of the Golden Ratio.
– Blue
Commented Sep 19, 2020 at 3:43
• @Blue Thank you for this interesting reference. Commented Sep 19, 2020 at 9:00
• If you're wanting to avoid fore-knowledge of the Golden Ratio because you think it's difficult to obtain, consider Odom's argument: Let midpt segment $\overline{PQ}$ (say, of length $s$) of an equilateral triangle extend to meet the circumcircle at $P'$ and $Q'$, and let $|PP'|=|QQ'|=1$. Then $P$ separates a side of the triangle (a chord of the circumcircle) into lengths $s$ and $s$, and it also separates $\overline{P'Q'}$ (another chord) into lengths $1$ and $s+1$. By the chord-chord aspect of the Power of a Point theorem, we have $s\cdot s=1\cdot(s+1)$, so that $s$ is the Golden Ratio. Done!
– Blue
Commented Sep 19, 2020 at 13:59
• No not because of difficulty. Just looking for alternative solutions to get the golden ratio bit. I agree that this argument is the simplest way to do that (and solve the original problem). But planned out the above proof and surprisingly got stuck: I think should be able to prove the circumcenter part without resorting to algebra beforehand and winkling out $\phi$ Commented Sep 19, 2020 at 14:20
• In a few hours time, I can give you a synthetic proof showing that $H$ is the center of the circumcircle of the triangle $BGF$ without resorting to golden ratio. But I am not sure whether this is that alternative solutions you are looking for to get the golden ratio bit. Let me know your opinion, so that I can take it down, if it does not fulfill your quest.
– YNK
Commented Oct 2, 2020 at 18:23

$$\underline{\mathrm{Note}}$$

Since we need to draw quite a few auxiliary lines and mark even more number of angles and points, our diagrams are going to get confusingly crowded. We draw all auxiliary lines as dotted lines, so that OP can easily differentiate between auxiliary lines and the lines mentioned in the question. The added points are denoted by red letters. The size of angles, which followed directly from the problem description are marked green. The measures of all derived angles are given in black, while the angles, which we are striving to determine, are shown in red.

$$\underline{\mathrm{Step\space 1}}$$

Consider $$\mathrm{Fig.\space 1}$$. As shown in this diagram, we have extended the OP’s sketch by drawing a line and the circumcircles of the equilateral triangle $$BGD$$. To be precise, a perpendicular has been dropped from the vertex $$D$$ of the triangle $$BGD$$ to opposite side $$BG$$ to meet it at $$Q$$. Since $$BGD$$ is an equilateral triangle, $$DQ$$ is the perpendicular bisector of $$BG$$. $$DQ$$ intersect the circumcircle of the triangle $$DEF$$ at $$P$$. Our aim is to prove that both the center of the circumcircle of $$BGF$$ and the point $$H$$, which is given as the diametrically opposite point of $$F$$ in OP’s sketch, lies on $$DQ$$.

Since $$\measuredangle BGD = \measuredangle FDG$$, the two lines $$BG$$ and $$DF$$ are parallel to each other. Therefore, the line $$DQ$$, which is perpendicular to $$BG$$, is perpendicular to $$DF$$ as well. This makes $$\measuredangle FDQ$$ a right angle and the circular arc $$FDP$$ a semicircle. Hence, $$FP$$ is a diameter of the circumcircle of the triangle $$DEF$$. However, it is given that $$FH$$ is a diameter of the same circle. Therefore, the two points $$P$$ and $$H$$ are one and the same. With this we have shown that $$H$$ lies on $$DQ$$, which is the perpendicular bisector of $$BG$$.

Furthermore, since $$BG$$ is the common chord of the circumcircles of the two equilateral triangles $$BGF$$ and $$BGD$$, their individual centers lies on $$DQ$$.

$$\underline{\mathrm{Step\space 2}}$$

The source of $$\mathrm{Fig.\space 2}$$ is also the OP’s sketch. We extend the existing line $$FH$$ to meet the circumcircle of $$BGF$$ at $$M$$. We denote the intersection point between $$FH$$ and $$DE$$ as $$R$$. We add two more new points, $$N$$ and $$O$$, which are the circumcenters of the triangles $$DEF$$ and $$ABC$$ respectively. Let $$\measuredangle FCD = \psi$$. Our aim is to show that $$FM$$ is a diameter of the circumcircle of $$BGF$$.

The chord $$BF$$ of the circumcircle of $$ABC$$ subtends $$\measuredangle FOB$$ at the circumcenter of $$ABC$$, while subtending $$\measuredangle FCB$$ at its circumference. Therefore, $$\measuredangle FOB = 2\measuredangle FCB = 2\psi$$. Since its sides $$FO$$ and $$OB$$ are radii of the circumcircle of $$ABC$$, $$FOB$$ is an isosceles triangle. Therefore, we shall write, $$\measuredangle OBF = 90^0-\psi$$.

Now, pay your attention exclusively to the angles at point $$B$$. Since $$OB$$ is a radius of its circumcircle, it is also the angle bisector of vertex angle $$B$$ of $$ABC$$. Therefore, $$\measuredangle OBC = 30^o$$, and hence, $$\measuredangle CBF = \measuredangle OBF- \measuredangle OBC = 90^o-\psi-30^o=60^o-\psi.$$

We keep on chasing angles to obtain, $$\measuredangle FBG = \measuredangle CBG- \measuredangle CBF = 60^o – \left(60^o-\psi\right) = \psi.$$

Both $$\measuredangle FMG$$ and $$\measuredangle FBG$$ are subtended at the circumference of the circumcircle of $$BGF$$ by the same chord $$GF$$. Thus, we have, $$\measuredangle FMG = \measuredangle FBG = \psi.$$

It is given that $$FH$$ is a diameter of the circumcircle of the triangle $$DEF$$, which makes $$FH$$ the perpendicular bisector of $$DE$$, the opposite side of its vertex $$F$$. This means that $$\measuredangle GRM$$ is a right angle. Now, we can determine the $$\measuredangle MGR$$ of the triangle $$MGR$$ as shown below. $$\measuredangle MGR = 180^o - \left( \measuredangle GRM + \measuredangle FMG\right) = 180^o - \left(90^o+\psi\right) = 90^o - \psi$$

Let us now divert the focus of our investigation to the two triangles $$DGF$$ and $$DFC$$. They share a side, namely $$DF$$. Furthermore, their angles $$FDG$$ and $$CDF$$ are equal. Since $$BD$$ and $$BG$$ are two sides of the same equilateral triangle, namely $$BGD$$, $$DG = BD$$. $$D$$ is the midpoint of $$BC$$, hence, $$CD = BD$$. Therefore, $$CD = DG$$. According to $$\mathit{Euclid I.4}$$ (also known as SAS Theorem), these two triangles are congruent, which gives us $$\measuredangle DGF = \measuredangle FCD = \psi$$. Thus, we shall write, $$\measuredangle MGF = \measuredangle MGR + \measuredangle DGF = 90^o - \psi + \psi = 90^o,$$ which indicates that $$FM$$ (or the extended $$FH$$) is indeed a diameter of the circumcircle of $$BGF$$.

$$\underline{\mathrm{Conclusion}}$$

In $$\mathrm{Step\space 1}$$, we showed that point $$H$$ and the circumcenter of $$BGF$$ lies on the perpendicular bisector of its side $$BG$$, namely $$DQ$$.

In $$\mathrm{Step\space 2}$$, we made it clear that $$FM$$, which actually is the extended $$FH$$, is a diameter of the circumcenter of $$BGF$$.

Now, the circumcenter of $$BGF$$ must lie on the intersection point of $$DQ$$ and $$FM$$. Since the point $$H$$ is common for both these lines, it is the sought circumcenter of $$BFG$$.

• Lovely answer explained in painstaking detail! Have verified that all steps are correct. I must thank you for this answer and the meticulous effort behind writing it. Commented Oct 3, 2020 at 2:24

Here is a way to brute force the result:

Since $$HF$$ is diameter, $$\angle HDG = \angle HDB = 30^\circ$$.

Hence $$\triangle HDG \cong \triangle HDB$$ and $$BH = HG$$.

This was the easy part.

We have found that $$DF = \frac{\sqrt3}4(\sqrt5-1)R$$.

Then $$DH = \frac 14(\sqrt5-1)R$$ and $$HF = \frac12(\sqrt5-1)R$$.

We also have $$JG = \frac{\sqrt3}4 R$$ and $$DJ = \frac 34R$$.

Finally finish off with Pythagoreas and many surd manipulations.

• Updated the question. Looking for some other solution to not use up this ratio already but arrive at it. Something without the brute force, the raison d'être of this exploration. But thank you for your answer! Commented Sep 19, 2020 at 10:36