# Golden Ratio in Equilateral and Right Isosceles Triangles

The following construction of the Golden Ratio has been posted by Tran Quang Hung at the CutTheKnotMath facebook page.

Let $ABC\;$ be an equilateral triangle inscribed in circle $(O).\;$ $D\;$ is reflection of $A\;$ through $BC.\;$ $MN\;$ is the diameter of $(O)\;$ parallel to $BC.\;$ $AD\;$ meets $(O)\;$ again at $P.$

Then, then circle centered at $D\;$ and passing through $B,C\;$ divides $PM,PN\;$ in the Golden Ratio.

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For a proof, assume $OB=1.\;$ Then $BC=CD=\sqrt{3},\;$ $DP=1,\;$ $NP=\sqrt{2}.\;$ In $\Delta DLP,\;$ $\angle DPL=135^{\circ}\;$ so that, with $PL=x,\;$ the Law of Cosines gives $x^2+1^2+\sqrt{2}x=(\sqrt{3})^2,\;$ i.e., $x^2+\sqrt{2}x-2=0.\;$ The equation has one positive root: $\displaystyle PL=x=\frac{\sqrt{10}-\sqrt{2}}{2}.\;$ From here, $\displaystyle NL=\sqrt{2}-\frac{\sqrt{10}-\sqrt{2}}{2}=\frac{3\sqrt{2}-\sqrt{10}}{2}.\;$ It follows that

$\displaystyle\frac{PL}{NL}=\frac{\sqrt{10}-\sqrt{2}}{2}\cdot\frac{2}{3\sqrt{2}-\sqrt{10}}=\frac{\sqrt{5}-1}{3-\sqrt{5}}=\varphi.$

### Golden Ratio

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