SOLUTION: the midpoint of the hypotnuse of a right triangle is equidistant from the vertices.
given: angle MON is a right angle, P is the midpoint of MN
Prove: MP= PN =OP
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-> SOLUTION: the midpoint of the hypotnuse of a right triangle is equidistant from the vertices.
given: angle MON is a right angle, P is the midpoint of MN
Prove: MP= PN =OP
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Question 31474This question is from textbook new york math a/b
: the midpoint of the hypotnuse of a right triangle is equidistant from the vertices.
given: angle MON is a right angle, P is the midpoint of MN
Prove: MP= PN =OP This question is from textbook new york math a/b
You can put this solution on YOUR website! the midpoint of the hypotnuse of a right triangle is equidistant from the vertices.
WHAT IS THE TRIANGLE NAME?MNO?
given: angle MON is a right angle,
OK...ASSUMING MNO IS THE TRIANGLE
MN IS HYPOTENUSE.
P is the midpoint of MN
SO MP=PN.
NOW IF WE TAKE MN AS DIAMETER AND P ITS MIDPOINT AS CENTRE AND DRAW A CIRCLE
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.
Prove: MP= PN =OP
MN IS
