a: \(\widehat{BAD}+\widehat{CAD}=90^0\)
\(\widehat{BDA}+\widehat{HAD}=90^0\)
mà \(\widehat{CAD}=\widehat{HAD}\)
nên \(\widehat{BAD}=\widehat{BDA}\)
b: \(\widehat{B}=90^0-40^0=50^0\)
\(\widehat{BDA}=\dfrac{180^0-50^0}{2}=65^0\)
\(\widehat{DAC}=90^0-65^0=25^0\)