Please help me out!!!!

Dear student,



ABCD is a rhombus. So all of its sides are equal.
Now, BC = DC 
⇒ ∠BDC = ∠DBC = x^o   (Angles opposite to equal sides are equal)
Also, ∠BCD = 60^o
∴ x^o + x^o + 60^o = 180^o 
⇒​ 2x^o = 120^o
⇒​ x^o = 60^o
i.e., ∠BCD = ∠BDC = ∠DBC =  60^o
So, ​∆BCD is an equilateral triangle.
∴ BD = BC = a
Also, OB = a /2
Now, in ∆OAB, we have:
 $A{B}^2= O{A}^2 + O{B}^2$ 
$$\begin{gathered} \Rightarrow O{A}^2=A{B}^2-O{B}^2=a^2-{\left({\displaystyle \frac{a}{2}}\right)}^2=a^2-\frac{a^2}{4}=\frac{3a^2}{4} \\ \Rightarrow O{A}^2=\frac{3a^2}{4} \\ \Rightarrow OA=\sqrt{\frac{3a^2}{4}} \\ \Rightarrow OA=\frac{\sqrt{3}a}{2} \\ \mathrm{Now}, AC=2\times OA=2\times \frac{\sqrt{3}a}{2}=\sqrt{3}a\therefore AC:BD=\sqrt{3}a:a =\sqrt{3} :1 \end{gathered}$$

Regards