The three triangles are all isoceles so $\angle \mathrm{BDC}=\angle \mathrm{ACB}=\angle \mathrm{ABC}.$

$\angle \mathrm{BDC}=\angle \mathrm{BAC}+\angle \mathrm{DBA}=2\angle \mathrm{BAC}.$ $\angle \mathrm{BCD}+\angle \mathrm{ABC}+\angle \mathrm{BAC}=\angle \mathrm{BDC}+\angle \mathrm{BDC}+\angle \mathrm{BAC}={180}^o.$ $2\angle \mathrm{BAC}+2\angle \mathrm{BAC}+\angle \mathrm{BAC}={180}^o$ $5\angle \mathrm{BAC}={180}^o$ $\angle \mathrm{BAC}=\angle \mathrm{DBA}={36}^o,\angle \mathrm{BDA}={108}^o$ $\angle \mathrm{ABC}=\angle \mathrm{ACB}=\angle \mathrm{BDC}={72}^o$ and $\angle \mathrm{CBD}={36}^o.$

The triangles $\mathrm{ABC}$ and $\mathrm{BDC}$ have lengths $\mathrm{AB}=\mathrm{AC}=p$ and $\mathrm{BC}=\mathrm{BD}=\mathrm{DA}=q$ and the angles are ${36}^o,{72}^o$ and ${72}^o$ so they are similar triangles. Taking the ratio of corresponding sides $\mathrm{AC}/\mathrm{BC}=\mathrm{BC}/\mathrm{DC}$ :

${\displaystyle \frac{p}{q}=\frac{q}{p-q}}$

So $p^2-\mathrm{pq}-q^2=0$ and dividing by $q^2$ gives the quadratic equation

${\displaystyle (p/q{)}^2-(p/q)-1=0}$

which has the solutions $1/2(1\pm \sqrt{5})$ We don't want the negative root for such a ratio as it would make no sense. Hence $p/q=1/2(\sqrt{5}+1)$. Similarly $q/p=1/2(\sqrt{5}-1)$.

Triangles $\mathrm{ABC}$ and $\mathrm{BDC}$ are similar and the ratio of areas is the square of the ratio of corresponding sides. So area of
$\mathrm{BDC}:\mathrm{ABC}$ $=[1/2(\sqrt{5}-1{)}^2]:1$
$=1/4(6-2\sqrt{5}):1$
$=(3-\sqrt{5}):2$

So if the area of triangle $\mathrm{ABC}$ is 2 the area of triangle $\mathrm{BDC}$ is and the area of triangle $\mathrm{ABD}$ is $2-(3-\sqrt{5})=\sqrt{5}-1)$.