Ta có: $\widehat{BAD}+\widehat{ADB}+\widehat{DBA}=180^o$ (1)
$\widehat{CAD}+\widehat{ADC}+\widehat{DCA}=180^o$ (2)
Lấy (1) - (2) ta được: $\widehat{ADB}-\widehat{ADC}+\widehat{DBA}-\widehat{DCA}=0$
⇔$\widehat{ADB}-\widehat{ADC}+150^o=0$
Lại có:$\widehat{ADB}+\widehat{ADC}=180^o$
⇒$\widehat{ADB}=15^o$