New Golden Ratio Construct with Geogebra using Square and Triangle with Same Base Width. Geometric proof of golden section?

Question

The below construct of the golden ratio, based on the ratio of segment c to segemnt b, is so very close to PHI. Geogebra gives the value of 1.61957 instead of 1.61803.

Might anyone have any insight as to a geometric proof as to whether the ratio of c to b really is the golden ratio PHI?

The square and equilateral triangle have the same base width. One corner of the triangle is at the very center of the square, as drawn below. The line dividing c and b is drawn from the top vertex of the the triangle to the midpoint of the side of the square.

Thanks for your assistance!

Answer 1

A different approach.

The line that segments $b$ and $c$ lie on is the following:

$$y=\frac{1}{3}x$$

The line that ${HJ}$ lies on is given by the following:

$$y=(1+\sqrt3)\left(x-\frac{5}{2}\right)$$

Setting the equations equal to one another to find the intersection point:

$$\frac{1}{3}x=(1+\sqrt3)x-\frac{5+5\sqrt3}{2}$$

$$\frac{5+5\sqrt3}{2}=\frac{2+3\sqrt3}{3}x$$

$$\frac{15+15\sqrt3}{4+6\sqrt3}=x$$

If the distance from the $y$-axis to this $x$-coordinate is $1/\phi$ that from the coordinate to $x=15/2$, then the length of $c$ is $\phi$ times that of $b$.

$$\frac{\frac{15}{2}-\frac{15+15\sqrt3}{4+6\sqrt3}}{\frac{15+15\sqrt3}{4+6\sqrt3}}=x$$

$$\frac{\frac{30+45\sqrt3-15-15\sqrt3}{4+6\sqrt3}}{\frac{15+15\sqrt3}{4+6\sqrt3}}=x$$

$$x=\frac{1+2\sqrt3}{1+\sqrt3}=1.634\approx\phi$$

I hope this was somewhat helpful.

Answer 2

Let $DB$ intersect $FG$ at point $K$. Note that since $F$ is the center of the square, $K$ is the midpoint of $FG$.

Also, since $FI:HK=1:\sqrt{3}$, this implies that $FK:JK=1:\sqrt{3}$.

Thus $JK=\frac{\sqrt{3}}{1+\sqrt{3}} \times \frac{FG}{2}$.

Thus $b:c=CI:JG=\frac{1}{2}:\frac{\sqrt{3}}{1+\sqrt{3}} \times \frac{1}{2}+\frac{1}{2}$, which is not the golden ratio.