góc B=góc C+40 độ
\(\widehat{A}=180^0-\widehat{B}-\widehat{C}=180^0-\widehat{C}-40^0-\widehat{C}=140^0-2\cdot\widehat{C}\)
=>\(\widehat{BAD}=\widehat{CAD}=70^0-\widehat{C}\)
\(\widehat{ADB}=180^0-\widehat{C}-40^0-70^0+\widehat{C}=70^0\)
=>góc ADC=110 độ