We can prove that ABBC is the golden ratio

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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