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

ÐBDC = ÐBAC + ÐDBA = 2 ÐBAC. 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 1/2 (1 ±Ö5) We don't want the negative root for such a ratio as it would make no sense. Hence p/q = 1/2 (Ö5 + 1) . Similarly q/p = 1/2(Ö5 - 1) .

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 = [1/2(Ö5 - 1)² ] : 1
= 1/4(6 -2Ö5 ) : 1
= (3 - Ö5 ) : 2

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