∠A+∠ABD+∠CBD+∠C=180°
∠A=∠ABD
∠CBD=180°-∠C-∠BDC=180°-2∠C
∴∠A+∠A+180°-2∠C+∠C=180°
∴∠C=2∠A
∵∠ABC=∠C
∴∠A+∠ABC+∠C=180°
即∠A+2∠A+2∠A=180°
∠A=36°
∠A+∠ABD+∠CBD+∠C=180°
∠A=∠ABD
∠CBD=180°-∠C-∠BDC=180°-2∠C
∴∠A+∠A+180°-2∠C+∠C=180°
∴∠C=2∠A
∵∠ABC=∠C
∴∠A+∠ABC+∠C=180°
即∠A+2∠A+2∠A=180°
∠A=36°