document.write( "Question 174584: A triangle ABC has a median on side AC and AD which meets vertex B and C and the point where both median intersect named O. show that BOC =90˚ +BAC \n" ); document.write( "
Algebra.Com's Answer #129679 by jojo14344(1513)\"\" \"About 
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\n" ); document.write( "Good way to prove your theorem \"BOC=90˚+BAC\" is via Equilateral Triangle.
\n" ); document.write( "Let's see as per the conditions:
\n" ); document.write( "---->
\n" ); document.write( "RIGHT GRAPH: you see GREEN line from vertex B going to midpoint (median) of AC, and GREEN line from vertex C going to midpoint (median) of AB.\r
\n" ); document.write( "\n" ); document.write( "Evaluating:
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\n" ); document.write( "To PROVE: BOC =90˚ +BAC
\n" ); document.write( "As you see the values in the graph:
\n" ); document.write( "\"120%5Eo=90%5Eo%2B60%5Eo\"
\n" ); document.write( "120deg is not equal to 150deg
\n" ); document.write( "If we rewrite what to prove:--->BOC=60˚ +BAC, then,
\n" ); document.write( "\"120%5Eo=60%5Eo%2B60%5E0\"
\n" ); document.write( "\"120%5Eo=120%5Eo\", correct
\n" ); document.write( ".
\n" ); document.write( "*Note: if you increase the angle on BAC, BOC increases too.
\n" ); document.write( "Let's see if BAC=90˚:
\n" ); document.write( "----->
\n" ); document.write( ".Now, we prove: BOC=90˚ +BAC
\n" ); document.write( "BOC=90˚ + 90˚
\n" ); document.write( "BOC=180˚-----------> NOT TRUE, AS YOU SEE IN THE GRAPH.
\n" ); document.write( ".
\n" ); document.write( "So if you can double check the condition to prove please.
\n" ); document.write( "If you're not satisfied, try to poste it again.
\n" ); document.write( "Thank you,
\n" ); document.write( "Jojo\r
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