Xét ΔABC có BE là phân giác
nên AB/AE=BC/CE=2
=>AB=2AE
=>tan ABE=1/2
1+tan^2ABE=1/cos^2(ABE)
=>1/cos^2ABE=1+1/4=5/4
=>cos^2ABE=4/5
=>cos ABE=2/căn 5
cos ABC=cos (2*ABE)
\(=2\cdot\left(\dfrac{2}{\sqrt{5}}\right)^2-1=\dfrac{3}{5}\)
=>AB/BC=3/5
=>AB=6cm
=>AC=8cm