Question

Answer 1

\(\widehat{ADB}+\widehat{B}+\widehat{BAD}=180^0\)

\(\widehat{ADC}+\widehat{C}+\widehat{CAD}=180^0\)

mà \(\widehat{BAD}=\widehat{CAD}\)

nên \(\widehat{ADB}+\widehat{B}=\widehat{ADC}+\widehat{C}\)

mà \(\widehat{B}>\widehat{C}\left(AC>AB\right)\)

nên \(\widehat{ADB}< \widehat{ADC}\)