Golden Ratio in an Isosceles Trapezoid with a 60° Angle II

Proof 1

Let $AB=a,\,$ $BC=ar,\,$ and $CD=ar^2.\,$ Let the point $E\,$ be the orthogonal projection of $A\,$ on $CD.\,$ Then

$\displaystyle AE=\frac{ar\sqrt{3}}{2}\,\text{and}\,DE=\frac{ar^2-a}{2}.$

Applying the theorem of Pythagoras in $\Delta ADE\,$ one gets

$\displaystyle r^4-3r^2+1=0.$

Thus, $\displaystyle r^2=\frac{3+\sqrt{5}}{2}=\left(\frac{1+\sqrt{5}}{2}\right)^2\,$ and, consequently, $\displaystyle r=\frac{1+\sqrt{5}}{2}=\varphi,\,$ the Golden Ratio.

Proof 2

If $DE=x\,$ then $ar^2=a+2x.\,$ $x=ar\cdot\cos 60^{\circ}=\displaystyle \frac{1}{2}ar\,$ which gives the equation $ar^2=a+ar,\,$ i.e., $r^2-r-1=0.$

Acknowledgment

The statement and its proof have been kindly communicated to me by Dr. Luc Gheysens (Flanders, Belgium.) Proof 2 is by John Armstrong.

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