`a)` Xét $∆AHB$ và $∆CHA$ có:
`\qquad \hat{AHB}=\hat{CHA}=90°`
`\qquad \hat{ABH}=\hat{CAH}` (cùng phụ `\hat{BAH}`)
`=>∆AHB∽∆CHA` (g-g)
`=>{AH}/{CH}={HB}/{HA}`
`=>AH^2=CH.HB`
`=>h^2 =b' .c'` (đpcm)
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`b)` $∆ABC$ vuông tại $A$
`=>S_{∆ABC}=1/ 2 AC.AB`
Mà `S_{∆ABC}=1/ 2 BC. AH`
`=>1/2AC.AB=1/2BC.AH`
`=>AC.AB=BC.AH=>bc=ah` (đpcm)
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`c)` Ta có: $AB^2+AC^2=BC^2$ (định lý Pytago)
`\qquad 1/{AC^2}+1/{AB^2}`
`={AB^2 +AC^2}/{AC^2 .AB^2}`
`={BC^2}/{(AC.AB)^2}={BC^2}/{(BC.AH)^2}={BC^2}/{BC^2 .AH^2}=1/{AH^2}`
`=>1/{h^2}=1/{b^2}+1/{c^2}` (đpcm)