In isosceles triangle ABC, we have AB=AC=4. The altitude from B meets AC at H. If AH=2(HC) then determine BC.
A
2HC
4 H 4
B C
3HC = 4
HC = 4/3
AH = 8/3
BH^2 + AH^2 = AB^2
BH^2 (8/3)^2 = 4^2
BH^2 = 16 - 64/9 = [144 - 64 ] / 9 = 80 / 9
BH^2 + HC^2 = BC^2
80/9 + 16/9 = BC^2
96/9 = BC^2
sqrt (96) / 3 = (4/3)sqrt (6) = BC
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