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Interesting problem... I tried solving it through isosceles triangles, and I've proved it doesn't work on equilateral triangles. Can anyone give me some hints towards a geometric proof?

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I know it works, but I need a geometric proof that it is true. – Oliver Nov 2 '10 at 9:36
If you're going to take the coordinate geometry route, draw a line intersecting both the horizontal and vertical axes. If you know the slope of the line, you know the angles it makes with the horizontal and vertical axes. Reflect the line along the horizontal and vertical axes, and find the intersection point of those two reflected lines. – J. M. Nov 2 '10 at 9:37
(and if you actually tried out my suggestion, one look at the equations of those reflected lines will give you the logical conclusion, without having to solve for the intersection point) – J. M. Nov 2 '10 at 10:31

The reason you cannot find a geometric proof is because it is not true. Suppose you have a triangle ABC such that two angle bisectors are perpendicular, say the angle bisectors of A and B. Suppose they intersect at some point D, then we have a right-angled triangle ADB. If in ABC the angle at A is a and the angle at B is b, then in ADB the angle at A is a/2 and the angle at B is b/2. ADB right-angled triangle implies a/2+b/2=90, which implies a+b=180, contradiction.

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