document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=426510>Question 426510</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Prove: The midpoint of the hypotenuse of a right triangle is equidistant from the three vertices.\r<BR>\n" );
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document.write( "A(0.0)  B(a,0) C(0,b) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #296698 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=296698\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The midpoint of the hypotenuse BC is (<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=a%2F2 ALIGN=MIDDLE  ALT=\"a%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >, <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=b%2F2 ALIGN=MIDDLE  ALT=\"b%2F2\"   DISABLED_style= \"max-width:90%; height:auto;\" >). We can find the distance from each point to the midpoint and show that they are all equal.\r<BR>\n" );
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document.write( "A better way to prove this is to circumscribe a circle around ABC. Since BAC is a right angle, the hypotenuse BC is a diameter of the circle. We can draw the midpoint D and show that DA, DB, DC are all radii of the circle, hence they are equal. </SPAN>\n" );
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