a) Áp dụng định lý Pi-ta-go vào \(\Delta ABC\), ta có :
BC2= AB2+ AC2= 202+ 212= 400+ 441= 841(cm)
\(\Rightarrow\)BC= \(\sqrt{841}\)= 29(cm)
b) AD là phân giác Â
=> \(\dfrac{AB}{AC}\) = \(\dfrac{BD}{CD}\)
<=> \(\dfrac{AB}{BD}\) = \(\dfrac{AC}{CD}\) = \(\dfrac{AB+AC}{BD+CD}\) = \(\dfrac{41}{BC}\) = \(\dfrac{41}{29}\)
=> 29.AB = 41.BD
<=> BD = \(\dfrac{29.AB}{\text{41}}\) = \(\dfrac{29.20}{41}\)=\(\dfrac{580}{41}\)
DC= BC-BD= 29-\(\dfrac{580}{41}\)=\(\dfrac{609}{41}\)