`a)`
Ta có: `S=1/ 2 bh_b=1/ 2 ch_c=1/ 2 ah_a`
`=>b={2S}/{h_b}; c={2S}/{h_c};a={2S}/{h_a}`
`\qquad b + 2c = 3a`
`=> {2S}/{h_b} + 2.{2S}/{h_c} = 3.{2S}/{h_a}`
`=> 2S (1/{h_b} + 2/{h_c}) =2S . 3/{h_a}`
`=> 1/{h_b} + 2/{h_c} = 3/{h_a}` (đpcm)
$\\$
`b)` $ac=b^2$
Ta có:
`\qquad S=1/ 2 ab sinC=1/ 2 bc sinA=1/ 2 ac sinB`
`=>ab sinC=bc sinA=ac sinB`
`=>ab sinC. bc sinA=(ac sinB)^2`
`=>(acb^2) sinC.sinA=(ac)^2 sin^2 B`
`=>b^2.b^2 sinC.sinA=(b^2)^2 sin^2 B`
`=>sinC. sinA=sin^2 B` (đpcm)
