a: \(\widehat{ADB}+\widehat{ABD}=180^0-\widehat{BAD}\)
\(\widehat{ADC}+\widehat{ACD}=180^0-\widehat{CAD}\)
mà \(\widehat{BAD}=\widehat{CAD}\)
nên \(\widehat{ADB}+\widehat{ABD}=\widehat{ADC}+\widehat{ACD}\)
mà \(\widehat{ABD}>\widehat{ACD}\)
nên \(\widehat{ADB}< \widehat{ADC}\)
b: Xét ΔABC có AD là đường phân giác
nên AB/BD=AC/CD
mà AB<AC
nên BD<DC