# Golden Ratio and the Egyptian Triangle

The *golden ratio* is related to the ubiquitous *Egyptian* - triangle [Huntley, pp. 43-44].

Let ABC be such a triangle with BC = 3,

Indeed, BO being an angle bisector, O divides AC in the ratio of the sides

AO / CO = AB / BC = 5/3.

From here, AO = 5/2 and CO = 3/2. Thus the circle's radius r is 3/2. By the Power of a Point Theorem,

BP·BQ = BC^{2}.

In other words,

(BO - 3/2)·(BO + 3/2) = 3^{2}.

From which, BO = 3√5/2. We thus find BP = 3(√5 - 1)/2. And finally,

PQ / BP | = 2·r / [3(√5 - 1)/2] |

= 2 / (√5 - 1) | |

= 2 · (√5 + 1) / 4 | |

= (√5 + 1) / 2 = φ. |

(Incidently, the circle is tangent to the hypotenuse AB.)

### References

- H. E. Huntley,
*The Divine Proportion*, Dover, 1970

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