Xét `\triangle ABH` và `\triangle CBA` có:
`\hat{AHB}=\hat{CAB}(=90^\circ)`
`\hat{ABC}:\text{góc chung}`
$⇒ \triangle ABH \backsim \triangle CBA (\text{g-g})$
`=> (AB)/(CB)=(BH)/(BA) <=> AB^2=BH.BC`
Xét `\triangle ACH` và `\triangle BCA` có:
`\hat{AHC}=\hat{BAC}(=90^\circ)`
`\hat{ACB}:\text{góc chung}`
$⇒ \triangle ACH \backsim \triangle BCA (\text{g-g})(1)$
`(1)=> (AC)/(BC)=(CH)/(CA) <=> AC^2=CH.CB`
`(1)=>(AC)/(BC)=(AH)/(BA) <=> AC.AB=BC.AH`
Do $\triangle ABH \backsim \triangle CBA \text{và} \triangle ACH \backsim \triangle BCA$
$⇒ \triangle ABH \backsim \triangle CAH$
`=> (AH)/(CH)=(BH)/(AH) <=>AH^2=BH.CH`
Có: `AB.AC=BC.AH`
`<=> AH=(AB.AC)/(BC)`
`<=> 1/(AH)=(BC)/(AB.AC)`
`<=> 1/(AH^2)=(BC^2)/(AB^2. AC^2)`
`<=> 1/(AH^2)=(AB^2+AC^2)/(AB^2. AC^2)`
`<=> 1/(AH^2)=1/(AB^2)+1/(AC^2)`