Midpoint of hypotenuse of right triangle is circumcenter.
- Author:
- benquintana
Show that the midpoint of the hypotenuse of a right triangle is the circumcenter.
Proof #1:
We have right triangle ABC.
We have A at (0,0); B at (x,0); and C at (0,y)
Define D as the mid point of the hypotenuse. Therefore it is at (x/2, y/2).
Show AD = DB = DC
AD^2 = (y/2)^2 + (x/2)^2
AD^2 = 1/4(x^2 + y^2)
DB^2 = EB^2 + DE^2
DB^2 = (x - x/2)^2 + (y/2)^2
DB^2 = (x/2)^2 + (y/2)^2
DB^2 = 1/4 (x^2 + y^2)
Since D is the midpoint, DB = CD.
DB^2 = AD^2
Therefore DB = AD
We have, DB = CD, therefore AD = CD
QED
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Proof #2:
DE is the perpendicular bisector of AB.
All points on the perpendicular bisector are equidistant from the endpoints of the segment.
D is on the bisector...
Therefore A and B are both equidistant from D.
Therefore AD = BD = CD.
QED.