$$\eqalign{
& a)\,\,AH = \sqrt {A{C^2} - H{C^2}} = \sqrt {5{a^2} - {a^2}} = 2a. \cr
& \Rightarrow {1 \over {H{I^2}}} = {1 \over {H{A^2}}} + {1 \over {H{C^2}}} = {1 \over {4{a^2}}} + {1 \over {{a^2}}} = {5 \over {4{a^2}}} \cr
& \Rightarrow HI = {{2a} \over {\sqrt 5 }} \cr
& b)\,\,Xet\,\,{\Delta _v}AHC: \cr
& IC.AC = H{C^2} \cr
& Do\,\,AC = AB,\,\,HC = BH \cr
& \Rightarrow IC.AB = B{H^2} \cr
& c)\,\,Ap\,\,dung\,\,dinh\,\,li\,\,Ta - let: \cr
& {{HI} \over {BK}} = {{HC} \over {BC}} = {1 \over 2} \Rightarrow BK = 2HI \Rightarrow B{K^2} = 4H{I^2} \cr
& {1 \over {H{I^2}}} = {1 \over {H{A^2}}} + {1 \over {H{C^2}}} \cr
& \Leftrightarrow {4 \over {B{K^2}}} = {1 \over {A{H^2}}} + {1 \over {{1 \over 4}B{C^2}}} \cr
& \Leftrightarrow {4 \over {B{K^2}}} = {1 \over {A{H^2}}} + {4 \over {B{C^2}}} \cr
& \Leftrightarrow {1 \over {B{K^2}}} = {1 \over {B{C^2}}} + {1 \over {4A{H^2}}} \cr} $$
