Ngô Tấn Đạt
17/09/2017 at 12:01-
We have :
AH . BC = AB . AC (equals two \(S_{\Delta ABC}\))
So 2.AH.BC = 2.AB.AC (1)
Apply Pythagoras theorem , we have :
AB2 + AC2 = BC2 (2)
But (AH + BC)2 = AH2 + BC2 + 2.AH.BC (3)
(AB + AC)2 = AB2 + AC2 + 2AB.AC (4)
From (1) ; (2) ; (3) ; (4)
=> AH + BC > AB + AC
Selected by MathYouLike -
Faded 28/01/2018 at 20:43
We have :
AH . BC = AB . AC (equals two SΔABC
)
So 2.AH.BC = 2.AB.AC (1)
Apply Pythagoras theorem , we have :
AB2 + AC2 = BC2 (2)
But (AH + BC)2 = AH2 + BC2 + 2.AH.BC (3)
(AB + AC)2 = AB2 + AC2 + 2AB.AC (4)
From (1) ; (2) ; (3) ; (4)
=> AH + BC > AB + AC