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)![]() ![]() ![]() You can put this solution on YOUR website! \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: \n" ); document.write( " \n" ); document.write( "To PROVE: BOC =90˚ +BAC \n" ); document.write( "As you see the values in the graph: \n" ); document.write( " \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( " \n" ); document.write( " \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 \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |