You may use pencil, pens, ruler, compass, protractor and scissors on this test. You may not use a calculator, book or notes. Please take your time on each problem and show your work in the space provided. If you get stuck on a problem, move on to the next and return when you have time.
Lemma 3.3.2: We can construct the perpendicular bisector of any arbitrary line segment AB.
Proof: Construct CA(|AB|) and CB(|AB|); construct the two points of intersection D and D' of these circles.
Construct the intersection of line ST with segment AB; call this intersection point M.
Angle AMS and angle BMS are congruent, complementary angles; therefore they must be right angles.
Angle ASM and angle BSM are congruent because ray ST is an angle bisector.
Segment SM is congruent to itself.
By Angle-Side-Angle, triangle AMS is congruent to triangle BMS.
Segment AM is congruent to segment BM because corresponding parts of congruent triangles are congruent.
Therefore, S must be the midpoint of segment AB.//