document.write( "Question 388219: another one...\r
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Algebra.Com's Answer #275137 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "First we construct a triangle ABC whose sides are the medians,\r\n" );
document.write( "with AC = 6, BC = 8, and AB = 10\r\n" );
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document.write( "Triangle ABC happens to be a right triangle, since 6² + 8² = 10², so\r\n" );
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document.write( "angle CBA has measure arccos(\"6%2F10\") = arccos(.6) = arcsin(8/10)=arcsin(.8).\r\n" );
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document.write( "We will let AB be an actual median of the final triangle we are \r\n" );
document.write( "going to create.\r\n" );
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document.write( "We know that the three medians intersect at a point which divides\r\n" );
document.write( "each median into two parts such that the shorter part is \"1%2F3\" \r\n" );
document.write( "of that median, and, of course, the longer part is \"2%2F3\" of that\r\n" );
document.write( "median.   \r\n" );
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document.write( "So we locate point D such that AD is \"1%2F3\" of AB and of course,\r\n" );
document.write( "BD is \"2%2F3\" of AB.  So AD = \"10%2F3\" and BD = \"20%2F3\"\r\n" );
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document.write( "Point D is where all three medians of our final triangle will intersect.\r\n" );
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document.write( "Next we must draw a median-to-be EF of the final triangle through D parallel\r\n" );
document.write( "and equal to AC such that point D divides median-to be EF such that DF is\r\n" );
document.write( "\"1%2F3\" of median-to be EF and DE is \"2%2F3\" of median-to-be EF. Since\r\n" );
document.write( "EF is 6, we will draw DF 2 units long and DE 4 units long:\r\n" );
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document.write( "Next we will draw line segment EG through A so that AE = AG. \r\n" );
document.write( "AG will be a side of the final triangle, and A will be the midpoint\r\n" );
document.write( "of side EG of the final triangle.\r\n" );
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document.write( "Now we can calculate AE from triangle ADE by the law of cosines\r\n" );
document.write( "since we have Side-Angle-Side  We know that \r\n" );
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document.write( "side AD = \"10%2F3\", \r\n" );
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document.write( "angle ADE = angle EAB because they are alternate interior angles formed\r\n" );
document.write( "by transversal AB cutting parallel lines AC and EF.\r\n" );
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document.write( "So angle ADE = arccos(.6)\r\n" );
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document.write( "side DE = 4\r\n" );
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document.write( "\"AE%5E2=AD%5E2%2BDE%5E2-2AD%2ADE%2Acos%28ADE%29\"\r\n" );
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document.write( "\"AE%5E2=%2810%2F3%29%5E2%2B4%5E2-2%2810%2F3%29%284%29cos%28arccos%28.6%29%29\"\r\n" );
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document.write( "\"AE%5E2=%28100%2F9%29%2B16-%2880%2F3%29%28.6%29\"\r\n" );
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document.write( "\"AE%5E2=%28100%2F9%29%2B16-16\"\r\n" );
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document.write( "\"AE%5E2=100%2F9\"\r\n" );
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document.write( "\"AE+=+sqrt%28100%2F9%29\"\r\n" );
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document.write( "\"AE+=+10%2F3\"\r\n" );
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document.write( "And since A is the midpoint of EG,\r\n" );
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document.write( "side EG of the final triangle is twice that or \r\n" );
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document.write( "side EG = \"20%2F3\".  That's one of the sides of the final triangle.\r\n" );
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document.write( "Now if we have drawn everything right so far, points G, F, and B\r\n" );
document.write( "should be colinear and GB should be a side of the final triangle,\r\n" );
document.write( "with F at its midpoint.\r\n" );
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document.write( "Next we must calculate FG.  But to do that we must calculate\r\n" );
document.write( "angle E.  \r\n" );
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document.write( "We will use the law of sines in triangle ADE.\r\n" );
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document.write( "\"AD%2Fsin%28E%29=AE%2Fsin%28ADE%29\"\r\n" );
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document.write( "So angle E = arcsin(.8) = arccos(.6)\r\n" );
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document.write( "Now we can calculate FG by the law of cosines, since we have\r\n" );
document.write( "side-angle-side.\r\n" );
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document.write( "EG = \"20%2F3\", angle E = arccos(.6), EF = 6\r\n" );
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document.write( "\"FG%5E2=EG%5E2%2BEF%5E2-2%2AEG%2AEF%2Acos%28E%29\"\r\n" );
document.write( "\"FG%5E2=%2820%2F3%29%5E2%2B6%5E2-2%2820%2F3%29%286%29cos%28arccos%28.6%29%29\"\r\n" );
document.write( "\"FG%5E2=400%2F9%2B36-80%2A.6\"\r\n" );
document.write( "\"FG%5E2=400%2F9%2B36-48\"\r\n" );
document.write( "\"FG%5E2=292%2F9\"\r\n" );
document.write( "\"FG=sqrt%28292%29%2F3=sqrt%284%2A73%29%2F3=%282sqrt%2873%29%29%2F3\"\r\n" );
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document.write( "And since F is the midpoint of BG,\r\n" );
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document.write( "side BG of the final triangle is twice that or \r\n" );
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document.write( "side BG = \"4sqrt%2873%29%2F3\".  That's another side of the final triangle.\r\n" );
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document.write( "There is just one more side to find, and that is BE, which we will now\r\n" );
document.write( "draw in:\r\n" );
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document.write( "We have now completed the final triangle BEG.  We only need side BE.\r\n" );
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document.write( "We can find it by the law of cosines since we have side-angle-side\r\n" );
document.write( "in triangle BDE.\r\n" );
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document.write( "Side DE = 4, \r\n" );
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document.write( "Angle BDE is supplementary to angle ADE and so its cosine is the\r\n" );
document.write( "negative of the cosine of angle ADE, so angle BDE = arccos(-.6)\r\n" );
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document.write( "Side BD = \"20%2F3\"\r\n" );
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document.write( "\"BE%5E2=DE%5E2%2BBD%5E2-DE%2ABD%2Acos%28BDE%29\"\r\n" );
document.write( "\"BE%5E2=4%5E2%2B%2820%2F3%29%5E2-2%2A4%2A%2820%2F3%29cos%28arccos%28-.6%29%29\" \r\n" );
document.write( "\"BE%5E2=16%2B%28400%2F9%29-%28160%2F3%29%28-.6%29\"\r\n" );
document.write( "\"BE%5E2=832%2F9\"\r\n" );
document.write( "\"BE=sqrt%28832%29%2F3=sqrt%2864%2A13%29%2F3=8sqrt%2813%29%2F3\"\r\n" );
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document.write( "So we have found all three sides of triangle BEG.  \r\n" );
document.write( "We didn't draw in the third median, but we don't need to.\r\n" );
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document.write( "Edwin
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