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°