Let's assume the triangle is ABC.
Given that AB = AC = 4.
Let BH be the altitude from B meeting AC at H.
Given that AH = 2(HC).
Let's assume HC = x, then AH = 2x.
By using Pythagoras theorem in triangle ABH, we have:
AB^2 = AH^2 + BH^2
4^2 = (2x)^2 + BH^2
16 = 4x^2 + BH^2
BH^2 = 16 - 4x^2 ---- (1)
Similarly, by using Pythagoras theorem in triangle AHC, we have:
AC^2 = AH^2 + HC^2
4^2 = (2x)^2 + x^2
16 = 4x^2 + x^2
16 = 5x^2
x^2 = 16/5
x = √(16/5) = 4/√5 = 4√5/5
By substituting the value of x in equation (1), we have:
BH^2 = 16 - 4(4√5/5)^2
BH^2 = 16 - 4(16/5)
BH^2 = 16 - 64/5
BH^2 = (80 - 64)/5
BH^2 = 16/5
BH = √(16/5) = 4/√5 = 4√5/5
Since BH = HC, the length of BC is equal to BH + HC.
BC = 4√5/5 + 4√5/5
BC = (4√5 + 4√5)/5
BC = (8√5)/5
BC = (8/5)√5
Therefore, the length of BC is (8/5)√5.
In isosceles triangle ABC, we have AB=AC=4. The altitude from B meets AC at H. If AH=2(HC) then determine BC.
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