Question

[Bông Y Hà]

Cho tam giác ABC có ba góc nhọn đường cao AH

a. Cm AB² +AC² =BC/2 + 2AM²

b.. AC²-AB² = 2BC . HM ( AC > AB)

Answer 1

[Mysterious Person]

a) ta có : \(AB^2+AC^2=2AH^2+BH^2+CH^2\)

\(=2AM^2-2HM^2+\left(BM-HM\right)^2+\left(CM+HM\right)^2\)

\(=2AM^2-2HM^2+BM^2-2BM.HM+HM^2+CM^2+2CM.HM+HM^2\)

\(=2AM^2+BC^2-2BM.CM=2AM^2+BC^2-\dfrac{2BC^2}{4}\)

\(=2AM^2+\dfrac{BC^2}{2}\left(đpcm\right)\)

b) ta có : \(AC^2-AB^2=AH^2+HC^2-BH^2-AH^2\)

\(=HC^2-BH^2=\left(CM+HM\right)^2-\left(BM-HM\right)^2\)

\(=CM^2+2CM.HM+HM^2-BM^2+2BM.HM-HM^2\)

\(=2HM\left(CM+BM\right)=2HM.BC\left(đpcm\right)\)