`a)` $ABCD$ là hình bình hành (gt)
`=>AD`//$BC$`=>AD`//$CK$ (vì $K\in BC$)
`\qquad AB`//$DC$`=>AN`//$DC$ (vì $N\in AB$)
$\\$
Xét $∆AMN$ có $AN$//$DC$
`=>{DM}/{MN}={CM}/{AM}` (định lý Talet)
$\\$
Xét $∆ADM$ có $AD$//$CK$
`=>{MK}/{DM}={CM}/{AM}` (định lý Talet)
$\\$
`=>{DM}/{MN}={MK}/{DM}`
`=>DM^2=MN.MK` (đpcm)
$\\$
`b)` Ta có:
`\qquad {DM}/{MN}={CM}/{AM}` (câu a)
`=>{MN}/{DM}={AM}/{CM}`
`=>{MN}/{DM}+1={AM}/{CM}+1`
`=>{MN+DM}/{DM}={AM+CM}/{CM}`
`=>{DN}/{DM}={AC}/{CM}`
`=>{DM}/{DN}={CM}/{AC}` $(1)$
$\\$
`\qquad {MK}/{DM}={CM}/{AM}` (câu a)
`=>{MK}/{DM}+1={CM}/{AM}+1`
`=>{MK+DM}/{DM}={CM+AM}/{AM}`
`=>{DK}/{DM}={AC}/{AM}`
`=>{DM}/{DK}={AM}/{AC}` $(2)$
$\\$
Từ `(1);(2)` ta có:
`\qquad {DM}/{DN}+{DM}/{DK}`
`={CM}/{AC}+{AM}/{AC}`
`={CM+AM}/{AC}={AC}/{AC}=1`
Vậy ` {DM}/{DN}+{DM}/{DK}=1` (đpcm)
