# The Golden Ratio in a Circle and Equilateral Triangle. Geometric/Trigonometric Proof?

Geogebra gave me 1.618. . . . for the following Golden Ratio construction shown below.

First off, has anyone seen anything similar to this construction?

Basically begin with an equilateral triangle. Inscribe a circle inside of it. Connect the two upper points where the circle is tangent to the triangle. Find the midpoint. Find the midpoint of the side of the newly defined smaller triangle atop. Draw a line from that midpoint through the first midpoint and on to the border of the circle. The blue and yellow line segments, thusly defined, are in the ratio of 1.618 to 1.

Might anyone know of a quick trigonometric or geometric proof? Thanks!! :)

Before choosing the simpler manner of depicting and drawing this golden ratio construction above, using just a triangle, circle, and segment, I had originally constructed it in a more complicated manner as shown here, using three tangent circles and a fourth circle drawn through and connecting the centers: and here:

which can be greatly simplified to this depiction, as also shown above:

By the "chord-chord" aspect of the Power of a Point Theorem, we have $$|\overline{AX}||\overline{XB}| = |\overline{CX}||\overline{XD}|\quad\to\quad p\cdot p = (p+q)\cdot q \quad\to\quad\frac{p}{q} = \frac{p+q}{p} = \frac{|\overline{CX}|}{|\overline{XE}|}$$

The relation between $p$ and $q$ says exactly that $\frac{p}{q}$ is the golden ratio, $1.618\dots$. $\square$

I can't say that I've seen this construction before, but I don't exactly have a catalog of such things in my head. The power-of-a-point connection seems (almost) "obvious" in retrospect, so it likely has been observed before. Still, it's pretty neat. (It's my favorite of your constructions so far.)

• Thanks so much Blue! And thanks for the geogebra tip! I'm improving at it day by day. Thanks again for all your expertise and insights. And the "Power of a Point Theorem!" – Astrophysics Math May 2 '16 at 16:18
• Actually, this result might be considered a variant of "Odom's Construction" noted in Wikipedia. (Certainly, my approach to the proof matches the one given in Wikipedia ---though I didn't realize it at the time (but, again, it's pretty "obvious")--- and this speaks to my claim that "[t]he power-of-a-point connection [...] likely has been observed before".) – Blue May 31 '16 at 8:35
• Thanks @Blue ! But too, I believe it is enough different from Odom's construction to merit the quality of an individual, original construction. If someone could directly draw Odom's construction overlain on mine or vice versa, I may be more convinced of any absolute similarity. And again, I trust your word and will continue to look at it more closely and ponder this. Finally, some aesthetics and notions of beauty come into play here, so there is a human element! – Astrophysics Math May 31 '16 at 16:19

Observe that $DXY$ is similar to $YOZ$

Implying $DX/XY=YO/OX$ i.e $(a+b)/a=a/b$

$OY$ is constructed similar to $OX$

Error in the image: $DX=a+b$ ,not $OD$

• Haven't you just written $OD = b = a+b$, therefore $a=0$? – user305860 May 31 '16 at 6:55
• Should have been DX – Rigel Jun 8 '16 at 17:26