`a)` Xét $∆ABH$ vuông tại $H$
`=>AH^2+HB^2=AB^2` (định lý Pytago)
`=>AH^2=AB^2-HB^2` $(1)$
$\\$
Xét $∆ACH$ vuông tại $H$
`=>AH^2+HC^2=AC^2` (định lý Pytago)
`=>AH^2=AC^2-HC^2` $(2)$
Từ `(1);(2)=>AB^2-HB^2=AC^2-HC^2`
`=>HC^2-HB^2=AC^2-AB^2` (đpcm)
$\\$
`b)` `AB=11;AC=15;BC=20`
Đặt `HC=x\quad (0<x<20)`
Ta có: `HB=BC-HC=20-x`
$\\$
Vì `HC^2-HB^2=AC^2-AB^2` (câu a)
`=>x^2-(20-x)^2=15^2-11^2`
`=>x^2-(400-40x+x^2)=104`
`=>x^2-400+40x-x^2=104`
`=>40x=504`
`=>x={63}/5` (thỏa mãn)
`=>HC={63}/5`
$\\$
`=>HB=20-x=20-{63}/5={37}/5`
$\\$
`AH^2=AC^2-HC^2` (câu a)
`=15^2-({63}/5)^2={1656}/{25}`
`=>AH=\sqrt{{1656}/{25}}={6\sqrt{46}}/5`
Vậy `HC={63}/5;HB={37}/5; AH={6\sqrt{46}}/5`
