Golden Ratio in Two Equilateral Triangles
Here a most elegant construction of the Golden Ratio posted by Tran Quang Hung (with a proof) at the CutTheKnotMath facebook page.
Construction
$ABCD$ is a rhombus with $2AC=BD;$ $(O)$ the inscribed circle; $E$ and $F$ the points of intersection of $(O)$ with $BD.$
Then $F$ divides $DE$ in the golden ratio.
Proof
Let $M$ be the point of tangency of $(O)$ with $AD.$
Triangles $OMD$ and $AOD$ are similar. It follows that $\displaystyle\frac{MD}{MO}=\frac{OD}{OA}=2$. Thus, $MD=2OM=EF.$ By a property of tangent, $DM^2=DF\cdot DE.$ From this, $EF^2=DF\cdot DE,$ or $\displaystyle\frac{FD}{FE}=\frac{EF}{ED},$ as required.
Golden Ratio
- Golden Ratio in Geometry
- Golden Ratio in an Irregular Pentagon
- Golden Ratio in a Irregular Pentagon II
- Inflection Points of Fourth Degree Polynomials
- Wythoff's Nim
- Inscribing a regular pentagon in a circle - and proving it
- Cosine of 36 degrees
- Continued Fractions
- Golden Window
- Golden Ratio and the Egyptian Triangle
- Golden Ratio by Compass Only
- Golden Ratio with a Rusty Compass
- From Equilateral Triangle and Square to Golden Ratio
- Golden Ratio and Midpoints
- Golden Section in Two Equilateral Triangles
- Golden Section in Two Equilateral Triangles, II
- Golden Ratio is Irrational
- Triangles with Sides in Geometric Progression
- Golden Ratio in Hexagon
- Golden Ratio in Equilateral Triangles
- Golden Ratio in Square
- Golden Ratio via van Obel's Theorem
- Golden Ratio in Circle - in Droves
- From 3 to Golden Ratio in Semicircle
- Another Golden Ratio in Semicircle
- Golden Ratio in Two Squares
- Golden Ratio in Two Equilateral Triangles
- Golden Ratio As a Mathematical Morsel
- Golden Ratio in Inscribed Equilateral Triangles
- Golden Ratio in a Rhombus
- Golden Ratio in Five Steps
- Between a Cross and a Square
- Four Golden Circles
- Golden Ratio in Mixtilinear Circles
- Golden Ratio in Isosceles Right Triangle, Square, and Semicircle
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