Giải thích các bước giải:
a.Xét $\Delta ABD, \Delta ADE$ có:
Chung $AD$
$\widehat{BAD}=\widehat{DAE}$
$AB=AE$
$\to\Delta ABD=\Delta AED(c.g.c)$
$\to DE=DB$
Vì $AB<AC\to \hat B>\hat C$
$\to \widehat{AED}=\widehat{ABD}=\widehat{ABC}>\widehat{ACB}=\widehat{ECD}$
Ta có $\widehat{DEC}=180^o-\widehat{AED}=180^o-\hat B=\hat A+\hat C>\hat C$
$\to \widehat{DEC}>\widehat{ECD}$
$\to DE<DC$
Mà $DE=DB\to BD<DC$
b.Vì $AB<AC, E\in $ cạnh $AC\to E, AE=AB$ nằm giữa $A, C$
$\to $ tia $DE$ nằm giữa tia $DA, DC$
$\to \widehat{ADC}=\widehat{ADE}+\widehat{EDC}>\widehat{ADE}=\widehat{ADB}$
