document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=139687>Question 139687</A>: <I><SPAN STYLE=\"word-break: break-all;\"> )    Use the fact that every point on the perpendicular bisector of a line segment is equally distant from the endpoints of the segment to find the center of the following rotation.\r<BR>\n" );
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document.write( " I can't get the picture to come up.  I figured out how to show it w/ a compass and straight edge but cannot figure out how to explain I know it's true that rotating ABC to A'B'C' can happen when you take segment A to A' and create the perpendicular bisector do the same for b and c and where all 3 bisectors intersect is the center. Why? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #101886 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=101886\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I think (and this is just my humble opinion, mind you) that you are making this way too difficult.\r<BR>\n" );
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document.write( "Given, line segment AB and line segment PQ that is the perpendicular bisector of AB.  Let Point X be the point of intersection of the two line segments.\r<BR>\n" );
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document.write( "Choose any point R on the segment PQ.  Construct segments RA and RB.\r<BR>\n" );
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document.write( "Then RX = RX by identity, AX = BX by the definition of a perpendicular bisector.  Angle RXA is a right angle and Angle RXB is a right angle by definition of perpendicular.  Two triangles with two equal sides and equal included angles are congruent by some theorem or another -- you look it up.  Therefore RA = RB, but RA is the distance from R to A and RB is the distance from R to B.  Therefore R is equidistant from A and B.  QED.\r<BR>\n" );
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