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document.write( "The link worked for me.\r\n" );
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document.write( "We know that triangle is a n isosceles right triangle, so angles A and C are\r\n" );
document.write( "both 45°. Since we know that triangle ABC is an isosceles right triangle, we\r\n" );
document.write( "can use ratio and proportion with the standard 45-45-90 right triangle to find\r\n" );
document.write( "BC:\r\n" );
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document.write( "We need to prove that triangles ANQ and BNP are congruent. We cannot just\r\n" );
document.write( "assume that they are. But all we have is SSA, which does not prove that two\r\n" );
document.write( "triangles are congruent. However, the SSA theorem is an \"either/or\" theorem:\r\n" );
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document.write( "The SSA theorem can be stated this way:\r\n" );
document.write( "If two sides and the non-included angle of one triangle are equal to the\r\n" );
document.write( "corresponding sides and angle of another triangle, the two triangles are either\r\n" );
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document.write( "1. congruent\r\n" );
document.write( "or\r\n" );
document.write( "2. the other non-included angles are supplementary.\r\n" );
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document.write( "So we must rule out the possibility that angles AQN and BPN are supplementary.\r\n" );
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document.write( "AN = NB because they are both 20. Angle NAQ = angle NBP and NQ = NP because they\r\n" );
document.write( "are sides of an equilateral triangle. So by the SSA theorem, either triangles\r\n" );
document.write( "ANQ and BNP are congruent or angles AQN and BPN are supplementary.\r\n" );
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document.write( "Let's let angle QPC have measure a. Let's put in the values of the angles at P and Q:\r\n" );
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document.write( "So angles AQN and BPN are not supplementary, because (30+a)+(120-a) equals\r\n" );
document.write( "150, not 180. Thus by the SSA theorem, triangles ANQ and BNP are congruent,\r\n" );
document.write( "and 30°+a = 120°-a\r\n" );
document.write( " 2a = 90°\r\n" );
document.write( " a = 45°\r\n" );
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document.write( "Next we will put in the actual numerical values for the angles at P and Q.\r\n" );
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document.write( "Now we can find BP for we have ASA in triangle BNP. We use the law of sines.\r\n" );
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document.write( "We know that 
and\r\n" );
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document.write( "Substituting:\r\n" );
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document.write( "Multiplying both sides by 4:\r\n" );
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document.write( "Rationalizing,\r\n" );
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document.write( "Now since PC = BC - BP and BC = 
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document.write( "Edwin
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document.write( "![\"\"](\"/images/stars/stars-13.gif\")
