In this diagram, the triangle ACD is isosceles, since the two sides, AC and AD, are equal as are the two angles ADC and ACD. [Also, angles ADC and ACD are twice angle CAD.]
If we bisect the base angle at D by a line from D to point B on AC then we have the angles as shown. BDC is then an isosceles triangle so CD=BD.
Since ABD is also isosceles, then DB=AB also, so CD=BD=AB.
We also note that triangles BCD and CDA are similar since their angles are equal. AB=CD so
which is the ratio of the lengths of the long side to the base in a 36°-72°-72° triangle. In the 36°-72°-72° degree triangle ADC, it is the same as the ratio of AC to CD, so:
We have shown that CD=AB so now
Putting these equalities together we have:
and we have called this ratio r.
If we let BC be of length 1, then we have AB=r (since AB/BC=r above) and AC=AB+BC=1+r, or: r/1=(1+r)/r, ie r2=1+r, the equation which defined the golden ratio (and a negative quantity, but lengths are positive).
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