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

$\angle \mathrm{BCD}+\angle \mathrm{ABC}+\angle \mathrm{BAC}=\angle \mathrm{BDC}+\angle \mathrm{BDC}+\angle \mathrm{BAC}={180}^o.$

$2\angle \mathrm{BAC}+2\angle \mathrm{BAC}+\angle \mathrm{BAC}={180}^o$
$5\angle \mathrm{BAC}={180}^o$,
$\angle \mathrm{BAC}=\angle \mathrm{DBA}={36}^o$,
$\angle \mathrm{BDA}={108}^o$ ,
$\angle \mathrm{ABC}=\angle \mathrm{ACB}=\angle \mathrm{BDC}={72}^o$
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 $(1\pm \sqrt{5})/2$. 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 $\mathrm{ABC}$ and $\mathrm{BDC}$ are similar and the ratio of areas is the square of the ratio of corresponding sides. So area of
$\mathrm{BDC}:\mathrm{ABC}$ $=[(\sqrt{5}-1{)}^2/2]:1$
$=(6-2\sqrt{5})/4:1$
$=(3-\sqrt{5}):2$

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