The triangles $ ABC $ and $ BDC $ have lengths $ AB = AC = p $
and $ BC = BD = DA = q $ and the angles are $ 36^o , 72^o $ and $
72^o $ so they are similar triangles. Taking the ratio of
corresponding sides $ AC/BC = BC/DC $ : $$\frac{p}{q} =
\frac{q}{p-q}$$ So $ p^2 -pq-q^2 = 0 $ and dividing by $ q^2 $
gives the quadratic equation $$(p/q)^2 -(p/q)-1=0 $$ which has the
solutions $(1 \pm \sqrt 5) /2$. We don't want the negative root for
such a ratio as it would make no sense. Hence $ p/q = (\sqrt{5} +
1)/2 $. Similarly $ q/p = (\sqrt 5 - 1)/2 $.
Triangles $ ABC $ and $ BDC $ are similar and the ratio of
areas is the square of the ratio of corresponding sides. So area
of
$ BDC : ABC $ $ = [(\sqrt 5 - 1)^2/2 ] : 1 $
$ = (6 -2\sqrt 5 )/4 : 1 $
$ = (3 - \sqrt 5 ) : 2 $
So if the area of triangle $ ABC $ is $2$ the area of triangle
$ BDC $ is $3-\sqrt 5$ and the area of triangle $ ABD $ is $ 2 - (3
- \sqrt 5) = \sqrt 5 -1 $.