Golden Ratio and the Egyptian Triangle

The golden ratio is related to the ubiquitous 3-4-5 - Egyptian - triangle [Huntley, pp. 43-44].

Let ABC be such a triangle with BC = 3, AC = 4 and AB = 5. Let O be the foot of the angle bisector at B. Draw a circle with center O and radius CO. Extend BO to meet the circle at Q and let P be the other point of intersection of BO with the circle. Then PQ / BP = φ.

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

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 = BC².

In other words,

(BO - 3/2)·(BO + 3/2) = 3².

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

Golden Ratio

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PQ / BP	= 2·r / [3(√5 - 1)/2]
	= 2 / (√5 - 1)
	= 2 · (√5 + 1) / 4
	= (√5 + 1) / 2 = φ.