+1167 In triangle ABC, angle ACB = 90 degrees. Let H be the foot of the altitude from C to side AB.
If BC = 1 and AH = 2, find CH.
+129771 A
2
H
C 1 B
Triangle AHC ≈ Triangle ACB
AH / CH = AC / BC
2 / CH = AC
CH = 2 / AC
Triangle BHC ≈ BCA
CH / BC = AC / BA
CH = AC / BA
2 /AC = AC / BA
2BA = AC^2
2BA + BC^2 = AB^2
2BA + 1^2 = AB^2
AB^2 - 2AB - 1 = 0
AB^2 - 2AB = 1
AB^2 -2AB + 1 = 2
(AB - 1)^2 = 2
AB - 1 = sqrt(2)
AB = 1 + sqrt (2)
BH = AB - AH = 1 + sqrt (2) - 2 = sqrt (2) - 1
CH^2 + BH^2 = BC^2
CH^2 + (sqrt (2) - 1)^2 = 1^2
CH^2 + (1 - 2sqrt 2 + 2) = 1
CH^2 = 2sqrt(2) - 2
CH = sqrt [ 2sqrt (2) - 2 ] ≈ .91
