From Gogeometry
Using :
- Let b=AD=DC and d=BC
- ∠BAC= ∠CBD and ∠BCD= ∠BCA
- => ΔCAB is similar to ΔCBD
- =>AC/BC=CB/CD=AB/BD => 2b^2=d^2, which means that d is the hypothenuse of a right triangle with side b
- Define C’ such as ΔDBC’ is the symmetry of ΔDBC by BD
- => ΔDBC’ is congruent to ΔDBC
- ∠BDC=180°-45°=135° = ∠BDC’ => ∠CDC’=360-135-135=90°
- => CC’=d and ΔCBC’ is equilateral
- ∠DBC= ∠DBC’= 2x=60°
- Therefore x=30°