document.write( "Question 1153572: In right triangle ABC, AC = BC and ∠C = 90. Let P and Q be points on hypotenuse
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\n" ); document.write( "to rotate it 90 degrees counter clockwise around C, let P go to P’ and find ∠P’CQ.
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Algebra.Com's Answer #775860 by AnlytcPhil(1806)\"\" \"About 
You can put this solution on YOUR website!
[My previous solution had some lettering that did not match the drawn figure.  I think I have corrected all the errors below.]\r\n" );
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document.write( "Draw lines CR and RB so that ΔCRB ≅ ΔCPA\r\n" );
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document.write( "Now draw in QR:\r\n" );
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document.write( "You can finish now.  It's mostly corresponding parts of\r\n" );
document.write( "congruent triangles and the Pythagorean theorem.  Here are \r\n" );
document.write( "some of the steps:\r\n" );
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document.write( "ΔCRB ≅ ΔCPA\r\n" );
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document.write( "BR = AP\r\n" );
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document.write( "∠ACP = ∠BCR  \r\n" );
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document.write( "∠ACP + ∠PCQ + ∠QCB = 90°\r\n" );
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document.write( "∠ACP + 45° + ∠QCB = 90°\r\n" );
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document.write( "∠ACP + ∠QCB = 45°\r\n" );
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document.write( "∠BCR + ∠QCB = 45°\r\n" );
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document.write( "Show that  ΔQCP ≅ ΔQCR\r\n" );
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document.write( "Then PQ = QR\r\n" );
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document.write( "≅QBR is a right triangle\r\n" );
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document.write( "BR² + BQ² = QR²\r\n" );
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document.write( "AP² + BQ² = PQ²\r\n" );
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document.write( "If you have trouble finishing, tell me in the space below,\r\n" );
document.write( "and I'll get back to you be email.  (No charge, ever!!)\r\n" );
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document.write( "Edwin
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