document.write( "Question 1025587: Let ABC be any triangle. Equilateral triangles BCX, ACY, and BAZ are constructed such that none of these triangles overlaps triangle ABC.\r
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\n" ); document.write( "\n" ); document.write( "b) Show that, regardless of choice of triangle ABC, we always have AX = BY = CZ\r
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Algebra.Com's Answer #641083 by ikleyn(52847)\"\" \"About 
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\n" ); document.write( "Let ABC be any triangle. Equilateral triangles BCX, ACY, and BAZ are constructed such that none of these triangles overlaps triangle ABC.\r
\n" ); document.write( "\n" ); document.write( "a) Draw a triangle ABC and then sketch the remainder of the figure. It will help if triangle ABC is not isosceles (or equilateral).\r
\n" ); document.write( "\n" ); document.write( "b) Show that, regardless of choice of triangle ABC, we always have AX = BY = CZ\r
\n" ); document.write( "\n" ); document.write( "Use much BASIC proofs as possible.
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\n" ); document.write( "\n" ); document.write( "Solution 1\r
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document.write( "Let us prove that  |AX| = |BY|.  For other combinations of segments the proofs will be the same.\r\n" );
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document.write( "From the triangle ACX, according to the cosines law\r\n" );
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document.write( "\"abs%28AX%29%5E2\" = \"abs%28AC%29%5E2+%2B+abs%28CX%29%5E2+-+2%2Aabs%28AC%29%2Aabs%28CX%29%2Acos%28ACX%29\".   (1)   ( argument ACX under cos is the angle ACX ).\r\n" );
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document.write( "Notice that LACX = LC + 60°,  where LC is the angle C of the original triangle ABC.\r\n" );
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document.write( "From the triangle BCY, according to the cosines law\r\n" );
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document.write( "\"abs%28BY%29%5E2\" = \"abs%28CY%29%5E2+%2B+abs%28BC%29%5E2+-+2%2Aabs%28CY%29%2Aabs%28BC%29%2Acos%28BCY%29\".   (2)   ( argument BCY under cos is the angle BCY ).\r\n" );
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document.write( "Notice that LBCY = LC + 60°,  where LC  is again the angle C of the original triangle ABC.\r\n" );
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document.write( "Now,  |AC| = |CY|,  |CX| = |BC|  and  LACX = LC + 60° = LBCY. \r\n" );
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document.write( "Therefore, right sides of (1) and (2) are equal.\r\n" );
document.write( "Hence, their left sides are equal.\r\n" );
document.write( "It implies that  |AX| = |BY|.\r\n" );
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document.write( "The statement is proved.\r\n" );
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document.write( "If you make a sketch, the proof will be crystally clear to you.\r\n" );
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document.write( "Again, let us prove that  |AX| = |BY|. \r\n" );
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document.write( "Consider triangles  ACX  and  BCY.\r\n" );
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document.write( "They have one pair of congruent sides CX  and  BC, and the other pair of congruent sides  AC  and  CY.\r\n" );
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document.write( "They also have congruent angles ACX and BCY that are concluded between the corresponding sides of these pairs.\r\n" );
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document.write( "Indeed, each angle ACX and BCT is the angle C of the original triangle ABC plus the angle of 60°.\r\n" );
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document.write( "So, the triangles  ACX  and  BCY  are congruent according with SAS-test for triangles congruency.\r\n" );
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document.write( "Thus  |AX| = |BY|, what has to be proved.\r\n" );
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document.write( "For other combinations of segments the proofs are very similar.\r\n" );
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document.write( "After the second proof, it becomes clear that if you rotate the triangle  ACX  around the vertex C in 60°, you will get the triangle  YCB. \r\n" );
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document.write( "It implies that the (directed) segments  AX  and  BY  make the angle of 120° (taking into account the directions of these segments).\r\n" );
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document.write( "So, the tree segments AX,  BY  and  CZ  make the angles of 120° between them.\r\n" );
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