# Golden Ratio via van Obel's Theorem

Here's another construction of the Golden Ratio by Bùi Quang Tuån.

The construction starts with the right isosceles triangle $ABC$ and is clear from the diagram:

In the diagram $\displaystyle\frac{OE}{EM}=\phi,$ the Golden Ratio. The following diagram may suggest why is this so.

With the reference to van Obel's theorem in $\Delta BCO,$

$\displaystyle\frac{OE}{EM}=\frac{OD}{DC}+\frac{OF}{FB}=2\frac{OD}{DC}.$

Assuming $OM=1=OD,$ $MC=2,$ $OC=\sqrt{2^{2}+1^{2}}=\sqrt{5},$ $DC=\sqrt{5}-1,$ so that

$\displaystyle\frac{OD}{DC}=\frac{1}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{4}.$

Thus, indeed $\displaystyle\frac{OE}{EM}=2\frac{\sqrt{5}+1}{4}=\frac{\sqrt{5}+1}{2}.$

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