b)\(\text{Gọi DE⊥AB}\)\(\text{→DE//AC}\)
Vì AD là tia phân giác của tam giác ABC
\(\Rightarrow BAD=DAC=\dfrac{1}{2}BAC=45^0\)
\(\Rightarrow EAD=45^0\)
\(\Rightarrow TamgiácAEDvuôngcântạiE\)
\(\rightarrow AD=AE\sqrt{2}\)
Mak AD là tia phân giác
\(\dfrac{\Rightarrow DB}{DC}=\dfrac{AB}{AC}=\dfrac{4}{3}\)
Mak\(\dfrac{DB}{DC}=\dfrac{EB}{AE}\left(địnhlýTalet\right)\)
\(\dfrac{\Rightarrow EB}{AE}=\dfrac{4}{3}\)
\(\Rightarrow\dfrac{AE}{AE+EB}=\dfrac{3}{7}\)
\(\Rightarrow\dfrac{AE}{AB}=\dfrac{3}{7}\Rightarrow AE=\dfrac{3}{7}.AB=\dfrac{12}{7}\)
\(\Rightarrow AD=AE.\sqrt{2}=\dfrac{12}{7}.\sqrt{2}=\dfrac{12\sqrt{2}}{7}\approx2,42\)
Xét tam giác ABC vuông tại A có AH đường cao
\(\Rightarrow AC^2=HC.BC\)
\(\Rightarrow BC=\dfrac{AC^2}{HC}=\dfrac{3^2}{1,8}=5\left(cm\right)\)
\(\Rightarrow HC=BC-HC=5-1,8=3,2\left(cm\right)\)
\(\Rightarrow AH^2=BH.HC\)
\(\Rightarrow AH^2=1,8.3,2=5,76\left(cm\right)\)
\(\Leftrightarrow AH=\sqrt{5,76}=2,4\left(cm\right)\)
\(\Rightarrow AB.AC=AH.BC\)
\(\Leftrightarrow AB=\dfrac{AH.BC}{AC}=\dfrac{2,4.5}{3}=4\left(cm\right)\)