document.write( "Question 31474This question is from textbook new york math a/b
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document.write( ": the midpoint of the hypotnuse of a right triangle is equidistant from the vertices.\r
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document.write( "given: angle MON is a right angle, P is the midpoint of MN\r
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document.write( "Prove: MP= PN =OP \n" );
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Algebra.Com's Answer #18149 by venugopalramana(3286)![]() ![]() You can put this solution on YOUR website! the midpoint of the hypotnuse of a right triangle is equidistant from the vertices. \n" ); document.write( "WHAT IS THE TRIANGLE NAME?MNO?\r \n" ); document.write( "\n" ); document.write( "given: angle MON is a right angle, \n" ); document.write( "OK...ASSUMING MNO IS THE TRIANGLE \n" ); document.write( "MN IS HYPOTENUSE. \n" ); document.write( " P is the midpoint of MN \n" ); document.write( "SO MP=PN. \n" ); document.write( "NOW IF WE TAKE MN AS DIAMETER AND P ITS MIDPOINT AS CENTRE AND DRAW A CIRCLE \n" ); document.write( "IT SHOULD PASS THROUGH O,SINCE ANGLE MON=90 AS GIVEN AND A SEMICIRCLE SUBTENDS 90 ANGLE AT ANY POINT ON ITS CIRCUMFERENCE AND VISE VERSA.AS O LIES ON CIRCLE AND PI NTHE CENTRE PO=RADIUS = DIAMETER/2=MN/2=MP=PN AS P IS MIDPOINT OF MN. \r \n" ); document.write( "\n" ); document.write( "Prove: MP= PN =OP\r \n" ); document.write( "\n" ); document.write( "MN IS \r \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "Prove: MP= PN =OP \n" ); document.write( " |