`a)` Xét $∆ABC$ và $∆HBA$ có:
`\qquad \hat{B}` chung
`\qquad \hat{BAC}=\hat{BHA}=90°`
`=>∆ABC∽∆HBA` (g-g) (đpcm)
`=>{AB}/{HB}={BC}/{BA}`
`=>AB^2=BH.BC` (đpcm)
$\\$
`\qquad {AC}/{HA}={BC}/{BA}`
`=>AB.AC=AH.BC` (đpcm)
$\\$
`b)` Xét $∆ABC$ và $∆HAC$ có:
`\qquad \hat{C}` chung
`\qquad \hat{BAC}=\hat{AHC}=90°`
`=>∆ABC∽∆HAC` (g-g)
`=>{AC}/{HC}={BC}/{AC}`
`=>AC^2=CH.CB` (đpcm)