Suppose we are given line segment AB. We then construct two circles X and Y -- X centered at A with radius AB and Y centered at B with radius BA. Let C and D be the intersections of X and Y. Let E be the point for which AB and CD intersect. We wish to prove that [math]\Delta[/math]ABC [math]\cong[/math] [math]\Delta[/math]ABD.[br][br]Note that by the definition of a circle, we know that AC [math]\cong[/math] AB. Similarly, we know that BA [math]\cong[/math] BC. Also, since AB [math]\cong[/math] BA, it follows by CN1 that AC [math]\cong[/math] BC. Therefore, we have that AC [math]\cong[/math] BC [math]\cong[/math] AB. By the definition of equilateral triangle, we have that ABC is an equilateral triangle. [br][br]Analogously, by the definition of circle, we know that AB [math]\cong[/math] AD and BA [math]\cong[/math] BD. Also, since AB [math]\cong[/math] BA, it follows by CN1 that AD [math]\cong[/math] BD. Therefore, AD [math]\cong[/math] BD [math]\cong[/math] AB. Once more, by the definition of equilateral triangle, we have that ABD is an equilateral triangle. Since both ABC and ABD contain side AB, by Proposition 8, we have that [math]\Delta[/math]ABC [math]\cong[/math] [math]\Delta[/math]ABD.