document.write( "Question 31474This question is from textbook new york math a/b
\n" ); document.write( ": the midpoint of the hypotnuse of a right triangle is equidistant from the vertices.\r
\n" ); document.write( "\n" ); document.write( "given: angle MON is a right angle, P is the midpoint of MN\r
\n" ); document.write( "\n" ); document.write( "Prove: MP= PN =OP
\n" ); document.write( "

Algebra.Com's Answer #18149 by venugopalramana(3286)\"\" \"About 
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( "
\n" );