+816 In isosceles triangle ABC, we have AB=AC=4. The altitude from B meets AC at H. If AH=2(HC) then determine BC.
+129895 A
H
B C
AH = (2/3) (4) = 8/3
HC = 4/3
Triangle BAH is right with BHA = 90
So
BH = sqrt (AB^2 - AH^2) = sqrt ( 4^2 - (8/3)^2) = sqrt [ 16 - 64/9 ] = sqrt [ (144 - 64) ] / 3 = sqrt (80 ) / 3
Triangle HBC is right with BHC = 90
BC =sqrt ( BH^2 + HC^2) = sqrt ( (sqrt (80) / 3)^2 + (4/3)^2 ) = sqrt [ 80/9 + 16/9] = sqrt ( 96) / 3 =
4sqrt (6) / 3
