lt is given that in $\Delta ABC$, $A B=B C$

and $\frac{AX}{XB}=\frac{AB}{AX}=\frac{1}{x}($ say $)$

$\Rightarrow AX=x·AB$

and $X B=x A X$

$\therefore XB=x^2\cdot AB$

$\because AB=AX+XB=x\cdot AB+x^2\cdot AB$

$\Rightarrow x^2+x-1=0$

$\Rightarrow x=\frac{-1\pm \sqrt{1+4}}{2}=\frac{\pm \sqrt{5}-1}{2}$

$\therefore$ $x>0,\text{ so }x=\frac{\sqrt{5}-1}{2}$ $\text{ Now, }\cos \theta =\frac{A{B}^2+B{C}^2-A{C}^2}{2\left(AB\right)\left(BC\right)}$ $\Rightarrow \cos \theta =\frac{2\left(A{B}^2\right)-\left(A{X}^2\right)}{2\left(A{B}^2\right)}[\because AB=BC]$ $=\frac{2\left(A{B}^2\right)-\left(x^2\cdot A{B}^2\right)}{2\left(A{B}^2\right)}$

$=\frac{2-x^2}{2}$

$=\frac{2-{\left(\frac{\sqrt{5}-1}{2}\right)}^2}{2}$

$=\frac{8-(5+1-2\sqrt{5})}{8}$

$=\frac{2\sqrt{5}+2}{8}$

$=\frac{\sqrt{5}+1}{4}=\cos {36}^{\circ }$

$\mathrm{So},\theta =\mathrm{\angle }ABC={36}^{\circ }$