tam giác$ AHC ~ BKC(g.g)\Rightarrow\frac{CK}{CH}=\frac{BK}{AH}\Leftrightarrow CK.AH=BK.CH$$\Rightarrow CK^2AH^2=BK^2.\frac{BC^2}{4}\Rightarrow (BC^2-BK^2).AH^2=BK^2.\frac{BC^2}{4}$
$\Rightarrow \frac{(BC^2-BK^2).AH^2}{AH^2.BK^2.BC^2}=\frac{BK^2.\frac{BC^2}{4}}{AH^2.BK^2.BC^2}\Rightarrow ĐPcm$