Math 3118, section 3
We're given the segment AB
Step 1. Draw (sufficiently large) circular arcs (of equal radius) centered at A and B respectively.
Step 2. Let C and D be the points where the two arcs intersect.
Draw the line that connects C and D. This line intersects the segment AB at the point M.
Step 3. We claim that (i) M is the midpoint of the segment AB, and (ii) the line CD is the perpendicular bisector of the segment AB.
Step 1. AC = BC because the two circles have equal radii. So, DACB is isosceles, and –ACB @ –ABC
Step 2. DACD @ DBCD by the SSS property. (Note that CD is congruent to itself.)
Since corresponding parts of congruent triangles are congruent, –ACD @ –BCD.
Step 3. DACM @ DBCM by the ASA postulate. (The point of steps 1 and 2 was just for checking the required angles for this step.) Since corresponding parts of congruent triangles are congruent, AM = BM, and –AMC @ –BMC. Since these two angles add to 180º, it follows that they are right angles.
Step 4 (conclusion). Since AM = BM, we conclude that M is the midpoint of AB.
Since –AMC and –BMC are right angles, the lines CD and AB are perpendicular.
Thus, we have constructed the perpendicular bisector.
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