`a)` Xét `\Delta ABH` và `\Delta ABC`, ta có:
`\hat{B} ` chung
`\hat{BHA}=\hat{BAC}=90^0`
`=>` `\Delta ABH `$\sim$` \Delta CBA` (g-g)
`=> \frac{AB}{CB}=\frac{BH}{AB}`
`=>AB^2=BC.BH` ( đpcm)`(1)`
`b)` Xét `\Delta AHC` và `\Delta` ` CBA`, ta có:
`\hat{C}` chung
`\hat{CHA}=\hat{CAB}=90^0`
`=>` `\Delta CBA `$\sim $`\Delta CAH` (g-g)`(2)`
`=> \frac{AC}{CH}=\frac{CB}{CA}`
`=>AC^2=CH.BC` (đpcm)
Từ `(1)` và `(2)` suy ra:
`\Delta CAH `$\sim$` \Delta ABH`
`=> \frac{AH}{BH}=\frac{CH}{AH}`
`=> AH^{2}=BH.CH`
`e)` Ta có:
`AH.BC=AB.AC`
`=> AH^2.BC^2=AB^2.AC^2`
`=> AH^2=\frac{AB^2.AC^2}{BC^2}`
`=>\frac{1}{AH^2}=\frac{BC^2}{AB^2.AC^2}`
`=> \frac{1}{AH^2}=\frac{AB^2+AC^2}{AB^2.AC^2}`
`=>\frac{1}{AH^2}=\frac{1}{AB^2}+\frac{1}{AC^2}`(đpcm)
