Hình thang EFCB có HE = HB, KF = KC[tex]\Rightarrow HK=\frac{1}{2}(EF+BC)[/tex]
Tam giác ABC có EB = AE, AF = FC[tex]\Rightarrow EF=\frac{1}{2}BC[/tex]
[tex]\Rightarrow HK=\frac{1}{2}(EF+BC)=\frac{1}{2}(\frac{1}{2}BC+BC)=\frac{1}{2}.\frac{3}{2}BC=\frac{3}{4}BC\Rightarrow BC=\frac{4}{3}HK[/tex]