Hint: If the area and the sides are rational, then the height is also going to be rational. Drop a perpendicular from A to BC, meeting the side at D. Now you only need to prove that the side is divided into two rational parts

If we call A,B,C the three sides of the traingle, and X,Y the two parts of the "devided" side of the triangle witch is C.
We will have X^2 + H^2 = A^2 & Y^2 + H^2 = B^2 => X^2 - Y^2 = A^2 - B^2 => X - Y = (A-B)(A+B)/( X+Y) = (A-B)(A+B)/C Or A-B , A+B, C are rational witch means (A-B)*(A+B)/C = X-Y is rational. C = X+ Y is rational => C + X - Y is also rational : C+ X-Y = 2X, 2X rational => X rational : X is rational and C is rational : C - X = Y is also rational. Hence X & Y are both rational.
You did the same ?

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TopNewestHint: If the area and the sides are rational, then the height is also going to be rational. Drop a perpendicular from A to BC, meeting the side at D. Now you only need to prove that the side is divided into two rational parts

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If we call A,B,C the three sides of the traingle, and X,Y the two parts of the "devided" side of the triangle witch is C. We will have X^2 + H^2 = A^2 & Y^2 + H^2 = B^2 => X^2 - Y^2 = A^2 - B^2 => X - Y = (A-B)

(A+B)/( X+Y) = (A-B)(A+B)/C Or A-B , A+B, C are rational witch means (A-B)*(A+B)/C = X-Y is rational. C = X+ Y is rational => C + X - Y is also rational : C+ X-Y = 2X, 2X rational => X rational : X is rational and C is rational : C - X = Y is also rational. Hence X & Y are both rational. You did the same ?Log in to reply

Nice use of contradiction. +1

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