Giải thích các bước giải:
\(\begin{array}{l}
a.Theo\;dinh\;ly\;Pytago:\\
A{D^2} = A{B^2} + B{D^2} \Rightarrow AD = \sqrt {{6^2} + {8^2}} = 10\\
b.\tan \widehat {BAD} = \frac{{BD}}{{BA}} = \frac{8}{6} = \frac{4}{3} \Rightarrow \widehat {BAD} \approx 53,{13^0}\\
\tan \widehat {BAC} = \frac{{BC}}{{BA}} = \frac{3}{6} = \frac{1}{2} \Rightarrow \widehat {BAC} \approx 26,{565^0}\\
\Rightarrow AC\;la\;phan\;giac\;\widehat {BAD}\\
c.DE//AB;AE \cap BD = C \Rightarrow \frac{{AB}}{{DE}} = \frac{{BC}}{{CD}} = \frac{3}{5} \Rightarrow DE = 10(cm)\\
\Rightarrow DE = AD \Rightarrow \Delta ADE\;can\;tai\;D\\
d.\Delta ADE\;can\;tai\;D \Rightarrow \widehat E = \widehat {DAE}\\
AB//DE \Rightarrow \widehat {BAC} = \widehat E(so\;le\;trong)\\
\Rightarrow \;\widehat {BAC} = \widehat {DAC} \Rightarrow AC\;la\;phan\;giac\;\widehat {BAD}
\end{array}\)