Xét `∆AHB` và `∆CAB` có:
`\hat{B}` chung
`\hat{AHB}=\hat{BAC}=90^0`
`⇒∆AHB~∆CAB` $(g.g)$
`⇒\frac{AB}{BH}=\frac{CB}{AB}`
`⇒AB^2=BH.CB.`
Xét `∆AHC` và `∆BAC` có:
`\hat{C}` chung
`\hat{AHC}=\hat{BAC}=90^0`
`⇒∆AHC~∆BAC` $(g.g)$
`⇒\frac{AC}{BC}=\frac{HC}{AC}`
`⇒AC^2=BC.HC.`
`⇒ \frac{AB^2}{AC^2}=\frac{BH.CB}{BC.HC}=\frac{HB}{HC}`
`⇒ (\frac{AB^2}{AC^2})^2=(\frac{HB}{HC})^2`
`⇔ \frac{AB^4}{AC^4}=\frac{HB^2}{HC^2}`
Xét `∆ECH` và `∆HCA` có:
`\hat{HCA}` chung
`\hat{AHC}=\hat{HEC}=90^0`
`⇒∆ECH∼∆HCA` $(g.g)$
`⇒\frac{EC}{HC}=\frac{CH}{CA}`
`⇒EC=\frac{CH.CH}{CA}`
`⇒EC=\frac{CH^2}{CA}.`
Xét `∆BDH` và `∆BHA` có:
`\hat{B}` chung
`\hat{BHA}=\hat{BDH}=90^0`
`⇒∆BDH∼∆BHA` $(g.g)$
`⇒\frac{BD}{BH}=\frac{BH}{BA}`
`⇒DB=\frac{BH.BH}{BA}`
`⇒DB=\frac{BH^2}{BA}.`
`⇒ \frac{DB}{EC}=\frac{BH^2}{BA}:\frac{CH^2}{CA}`
`⇔ \frac{DB}{EC}=\frac{BH^2}{BA}.\frac{CA}{CH^2}`
`⇔ \frac{DB}{EC}=\frac{BH^2}{CH^2}.\frac{CA}{BA}`
`⇔\frac{DB}{EC}=\frac{AB^4}{AC^4}.\frac{CA}{BA}`
`⇔\frac{DB}{EC}=\frac{AB^3}{AC^3}.`
Vậy ` (\frac{AB^2}{AC^2})^2=(\frac{HB}{HC})^2.`, `\frac{DB}{EC}=\frac{AB^3}{AC^3}.`
Tham khảo hình.
