The three triangles are all isoceles so ÐBDC = ÐACB = ÐABC.

ÐBCD + ÐABC + ÐBAC = ÐBDC + ÐBDC + ÐBAC = 180^o.

2 ÐBAC + 2 ÐBAC + ÐBAC = 180^o
5 ÐBAC = 180^o ,
ÐBAC = ÐDBA = 36^o,
ÐBDA = 108^o ,
ÐABC = ÐACB = ÐBDC = 72^o
and ÐCBD = 36^o.

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 :

p q = q p-q

So p² -pq-q² = 0 and dividing by q² gives the quadratic equation

(p/q)2 -(p/q)-1=0

which has the solutions (1 ±Ö5) /2. We don't want the negative root for such a ratio as it would make no sense. Hence p/q = (Ö5 + 1)/2 . Similarly q/p = (Ö5 - 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 = [(Ö5 - 1)²/2 ] : 1
= (6 -2Ö5 )/4 : 1
= (3 - Ö5 ) : 2

So if the area of triangle ABC is 2 the area of triangle BDC is 3-Ö5 and the area of triangle ABD is 2 - (3 - Ö5) = Ö5 -1 .