Đáp án:
`a.`Xét `\Delta \text{HBA}` và `\Delta \text{ABC}` có:
`\hat{\text{B}}` chung.
`\hat{\text{AHB}}=\hat{\text{BAC}}=90^@`
`=> \Delta \text{HBA} ∽ \Delta \text{ABC}` (g.g)
`b.`Xét `\Delta \text{HAC}` và `\Delta \text{ABC}` có:
`\hat{\text{C}}` chung.
`\hat{\text{AHC}}=\hat{\text{BAC}}=90^@`
`=> \Delta \text{HAC} ∽ \Delta \text{ABC}`(g.g)
`c.`Ta có: `\Delta \text{HBA} ∽ \Delta \text{ABC}`(cm c.a)
`=> \frac{\text{HB}}{\text{AB}}=\frac{\text{BA}}{\text{BC}}`(các cạnh tỉ lệ tương ứng)
`=> \text{HB.BC}=\text{AB.BA}`
`=> \text{HB.BC}=\text{AB}^2`
`=> \text{AB}^2=\text{BC.BH}`(đpcm)
