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H(#\\H$(H(FH\We draw angle BAC, then we draw a circle with A as center. The circle intersects the segmebts AB and AC at points E and F, respectively. We pick a point G at a convenient distance from F, then draw a circle with the center at F and radius FG. We also draw a circle with the center at E and the same radius. The two circles intersect in points H and I.
By construction, we have AE=AF and EH=FH. It follows that triangles AEH and AFH are congruent (SSS), so angles EAH and FAH are equal.
So, AH is the bisector of angle BAC.
Intuitively, there is no reason why this shouldn't work on the sphere: a (small) circle and a line segment (sector of a great circle) still intersect in 0, 1 or 2 points, and so do two (small) circles. Also, the circle definition guarantees that the trianlges sides are equal. The only "weak" link is whether SSS works on a sphere. t'ly.
e with A as center. The circle itersects the segmebts AB and AC at poi^MATH 131 Assignment 5 Adrian Ilie
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