document.write( "Question 1077820: In triangle ABC, the perpendiculars, AX BY and CZ drawn from the vertices A, B and C on the sides BC, CA and AB respectively meet at O. show that AO. OX = BO. OY = CO. OZ.
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Algebra.Com's Answer #692355 by ikleyn(52781) $\"\"$ $\"About$

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document.write( "0.  Make a sketch to follow my arguments.\r\n" );
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document.write( "1.  The triangles AOY and BOX are right-angled triangles.\r\n" );
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document.write( "    They have congruent acute angles AOY and BOX (these angles are congruent since they are vertical).\r\n" );
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document.write( "    Therefore, these triangles are SIMILAR.\r\n" );
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document.write( "    It implies that the corresponding sides are proportional:   = .\r\n" );
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document.write( "    Hence, |AO|*|OX| = |BO|*|OY|.\r\n" );
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document.write( "2.  The other equality is proved similarly. \r\n" );
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document.write( "    Prove it on your own as an exercise.  \r\n" );
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