\(\Delta\) ABC có AB = AC => \(\Delta\) ABC cân tại A
\(\widehat{A}\) = 1000 => \(\widehat{B}\) = \(\widehat{C}\) = \(\frac{180^0-100^0}{2}=40^0\)
\(\Delta\) ABD có BD = AB => \(\Delta\) ABD cân tại B
=> \(\widehat{A1}\) = \(\widehat{D1}\) = \(\frac{180^0-40^0}{2}=70^0\)
ta có \(\widehat{A1}\) + \(\widehat{A2}\) = 1000
mà \(\widehat{A1}\) = 700 => \(\widehat{A2}\) = 300