SOLUTION: ) 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
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Question 139687: ) 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.
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?
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
I think (and this is just my humble opinion, mind you) that you are making this way too difficult.
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.
Choose any point R on the segment PQ. Construct segments RA and RB.
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.
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