document.write( "Question 1201144: Given square ABCD, let P and Q be the points outside the square that make triangles
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document.write( "CDP and BCQ equilateral. Prove that triangle AP Q is also equilateral. \n" );
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Algebra.Com's Answer #835404 by math_tutor2020(3816)![]() ![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "Start with square ABCD. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Form equilateral triangles CDP and BCQ, such that the new points are outside the square. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Then form triangle APQ shown in red. \n" ); document.write( " \n" ); document.write( "The goal is to prove that triangle APQ is equilateral.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "For now let's focus on triangle ABQ. \n" ); document.write( "BQ = BC (since BCD is equilateral) \n" ); document.write( "AB = BC (since ABCD is a square) \n" ); document.write( "AB = BQ (by the transitive property)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "The AB = BQ statement then tells us triangle ABQ is isosceles. \n" ); document.write( "The vertex angle is 90+60 = 150 degrees. \n" ); document.write( "The base angles are (180-150)/2 = 30/2 = 15 degrees each.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Through similar logic, you'll find that triangles ADP and PCQ are congruent to isosceles triangle ABQ. \n" ); document.write( "All isosceles triangles mentioned have the same base length \n" ); document.write( "AQ = AP = PQ\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Therefore, triangle APQ is equilateral. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |