We know that:
∠ABC=∠BCA=xº
∠BDC=∠BCD=xº
1) we have to find the angle ∠DBC:
∠BDC+∠BCD+∠DBC=180º
xº+xº+∠DBC=180º
2xº+∠DBC=180º
∠DBC=180º-2xº
2) we have to find the angle ∠ABD:
∠ABD=∠ABC-∠DBC
∠ABD=xº-(180º-2x)
∠ABD=xº-180º+2xº
∠ABD=3xº-180º
Answer: ∠ABD=3xº-180º
