Many Isosceles

Let $\angle ABC = \theta$ .
Then $\angle BAD = \theta$ , $\angle ADC = 2 \theta$ and $\angle ACB = 2 \theta$ .
Then, $\angle DAC = 180^\circ - 4 \theta$ , and so
$2 \theta = \angle ACB = \angle CAB = \angle CAD + \angle DAB = (180^\circ - 4 \theta) + \theta$ .
This gives us $\theta = \frac{ 180^\circ } { 5 } = 36 ^ \circ$ .

Let $\angle ACB=\angle CAB=x.$
So $,\angle ADC=\angle ACD=x$ since $\triangle DAC$ is isosceles. $\angle DAC=180-2x.\angle ADB=180-x.$
So, since $\triangle BDA$ is isosceles, $\angle DBA=\angle DAB=\dfrac{x}{2}.\angle BAC=\angle BCA=x$ since $\triangle BAC$ is isosceles.
$\angle BAC=\angle BAD+\angle DAC\Rightarrow x=180-2x+\dfrac{x}{2}.$ After solving this, we get $x=72.$
Therefore, $\angle ABC=\dfrac{x}{2}=\dfrac{72}{2}=\boxed{36}.$

Let pink angle measures X.

<BAD=X, <ADC=2X, <ACD=2X, <CAB=2X,

<A+<B+<C=5X=180. X=180/5=36

As $AC=BD$ , we can clearly see that $BC=2AC$ . Now, as $\bigtriangleup\space ABC$ is isosceles, $AB=2AC=2BD=2DA$ . So, we have got a $\bigtriangleup\space ABD$ with sides in the ratio $2:1$ , with $AD=BD$ .

Hence, the angles of the $\bigtriangleup\space ABD$ are $x$ and $2x$ . So,

$x+2x+2x\space =\space 180\space \Rightarrow\space x\space =\space 36$ .